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Math 112 Quadratics- Notes

Resource: Mathematical Modeling Book 3

Instructional time frame for test #1

1 day of start up with introduction and handout of text books.8 days instruction1 day to test10 days to end of test

Assignments referred to as handouts are attached at the end of this document.

Note(1) Sequence- a series of numbers that are related in some way by a rule or pattern.e.g. 1,3,5,7,9,… the rule here is to add 2 to the previous term to get the next.

Note(2) Term- is a member or item in a sequence. We can represent a term using a letter and a subscript. Eg.t1 = term 1 and from the sequence above: t1 = 1 t2 = 3 t3 = 5 t4 = 7 t5 = 9

Note(3) General term- is represented by the letter t and the variable n so that:tn = term n or the nth term.

If n = 7 then tn = t7 If n = 8 then tn = t8

Note(4) Infinite sequence- a sequence that continues indefinitely.Notation: { t1 , t2 , t3 , t4 , t5 , …} ellipsis (3 dots) signifies continuance.

Note(5) Finite sequence- a sequence that terminates.Notation: { t1 , t2 , t3 , t4 , t5 , …, tn} tn signifies a termination

Asn(A1) Investigation #1 p. 2

Asn(A2) Answer questions 1,2,6 p. 3


Note(6) Arithmetic sequence- a pattern of numbers in which each term is found by adding or subtracting a number to (or from) the previous term.

e.g. 5, 7, 9, 11, … the next number is obtained by adding 2 to the previous term. It is, therefore, an arithmetic sequence.

Note(7) Common difference- if the differences between any two numbers in a sequence is always the same we say that we have a common difference.

e.g. 5, 7, 9, 11,… We have a common difference of +2

2 2 2

e.g. 1, 4, 9, 16… We do not have a common difference.

3 5 7

Note: If a sequence does not have a common difference (at the first level) it is not an arithmetic sequence.

Note(8) Arithmetic sequence formula:If t1 = the first term of a sequence and If d = common difference then we can construct a formula that can

be used for predicting any term in the sequence as follows:

tn = t1 + (n-1)d Where: t1 = the first term d = the common difference n = the term number (n-1) one less than the term number.

Asn(A3) Answer questions 7 to 12 p. 4,5 (Question 9 to 12 to be submitted for marking).


Note(9) Graphing properties of an arithmetic sequence- An arithmetic sequence has the following graphing properties:

a) it is linear- forms a straight lineb) the graph is discrete (no meaning between points)c) the slope of the line is equivalent to the common difference so that D1 = slope of the graph = d the common differenced) there is a constant rate of growth because the slope between any two points is the same.e) the domain of an arithmetic sequence is D= {x x N }

Asn(A4) A,B,C,D,E p. 7

Asn(A5) Q 17, 20 (Graph for Q 17 only) p. 8,9

Asn(A6) Q 22,23 p. 9 and Q 27,29 p. 10

Math 112 Quadratics- NotesNote(10) Quadratic sequence- is a sequence of numbers formed from an function containing a power of at least the second order. The general formula that generates a quadratic sequence is:

tn = an2 + bn + c Where: tn = term valuen = term numberD2 = 2a or a = D2

2Note(11) Properties of a quadratic sequence include:1) The common difference = D2

2) The multiplication of two arithmetic sequences generates a quadratic sequence. E.g. n(n-1) = n2 – n 3) In the equation tn = an2 + bn + c a = ½ D2 4) One of the variables must be of at least the order two.

Note(12) Graphing properties of a quadratic sequence:1) The graph of a quadratic sequence forms a parabola2) The slope of a quadratic is not a constant3) The graph is discrete. (not continuous)

Note(13) Power sequences- a sequence where one of the variables is of the second order or higher. Examples of power sequences and properties are:

Quadratic: tn = ax2 +bx +c1) Common difference at D2

2) D2 = 2a or a = D2

2Cubic: tn = ax3 +bx2 +cx +d

1) Common difference at D3

2) D3 = 6a or a = D3

6Quartic: tn = ax4 +bx3 +cx2 +dx +e

1) Common difference at D4

2) D4 =24a or a = D4

24Asn(A7) Handout on sequences.

Advanced. Handout on algebraic method. Q 38, 39, 40 p. 12,13


Note(14) Analyzing data- Given a table of values it is possible to determine the sequence of data as being either arithmetic or quadratic. First enter the data of the independent variable under L1 and the data for the dependent variable under L2. You may then determine the nature of the sequence by one of the steps given below:

1) Plot the data using: 2nd y= 1: Plot1 (choose the desired options) Set the Window parameters then press Graph. If the data forms a straight line then the relationship is arithmetic. If the data is curved or parabolic it is likely quadratic or one of the other power sequences (cubic, quartic, etc.)

OR

2) Determine the first and second differences and place results under L3 and L4 (if necessary). Place cursor at top of L3 and press: 2nd Stat Ops 7 List followed by name of the list for which you want differences taken. (Example L2) and repeat the procedure at the top of L4. If D1 is a constant the graph is arithmetic and if D2 is a constant the graph is quadratic

Note (15) Determining the line of best fit- Once you have determined whether the sequence of numbers is arithmetic or quadratic you may find the line of best fit by using linear regression or quadratic regression.

Linear regression: Stat Calc 4: LinReg EnterQuadratic regression: Stat Calc 5: QuadReg Enter

(For the golf ball question from Asn(A9) at the end of this document Quadregression gives a function y = -4.9x2 + 28.7x + 9)

Note(16) Graphing your line of best fit- The function generated by the calculator may be graphed by entering the function in your calculator.

1) Press Y=2) After \Y1 = (enter the function)3) Press Window and set the values for Xmin, Xmax, Ymin, Ymax4) Press Graph

Asn(A8) Q 8a) 5 a) b) c) 9 a) b) c) p. 16,17 and Q 17 p. 19

Math 112 Quadratics- NotesFor notes 17 to 21 refer to Q 1 of Asn(A9) – the flight of a golf ball.

Note(17) Determining the maximum or minimum value of a graph- The line of best fit for a quadratic always generates a curve that will have a maximum or minimum value. There are three ways of obtaining this maximum or minimum value:

1) Press Trace then move the “x” displayed on your graph left or right until it is centered over the max (or min) value. Read the coordinates for this point on the bottom of your screen.

2) Press 2nd Graph and read through the second column of Y values to find the highest (or lowest) value from the tables provided.

3) Press 2nd Trace then choose option 4: maximum . You will be required to set the Left and Right boundaries (pressing Enter each time) then press Enter a third time upon which the calculator will “look” between the two boundaries that you have set to locate the maximum (or minimum) value.

(2.9 , 51)Left boundary Right boundary

The boundaries must be pointing towards each other. This requires the left boundary be set before the right boundary. Pressing enter a third time engages the calculator to search for the maximum value of the graph.


Note(18) Axis of symmetry- A vertical line drawn through the vertex of the graph that defines the maximum (or minimum) value of a graph is called the axis of symmetry. The equation of the axis of symmetry equals the x value of the vertex. (2.9, 51).

The graph is split in two by this axis of symmetry so that the left side is a mirror image of the right side of the graph or we may say the graph is symmetrical about the line x = 2.9

Note(19) Maximum value of the graph- The maximum value of the graph is the y-coordinate in the vertex:

(2.9, 51 )

x = 2.9 is the equation of the axis of symmetry.

y = 51 is the maximum value for the graph.


Note(20) The Domain and Range of a graph- the domain and range of a graph can be determined from the table of values if one is given.e.g. From the golf ball question Asn(A9) at the end of these notes:

Time 1 2 3 4 5Height 32.8 46.8 51 45.4 30

The domain of the graph is the values that are determined by the independent variable, which in this case, is the time of the golf ball in flight. The lowest value here could be 0 seconds and the highest 6.2 seconds. The domain may be written two ways:

1) Interval notation x [0, 6.2]2) Set notation D = [ x 0 x 6.2, x R]

The range of a graph are the values determined by the independent variable and given to the dependent value by some function which in this case is the height of the golf ball. The lowest value here is 0 m as this is the height of the ball when it lands on the green and the highest value is 51 m as this is the maximum value of the ball in flight given by the vertex.The range may be written two ways:

1) Interval notation y [0,51]2) Set notation R= [y 0< y < 51, y R]


Note(21) Interpreting the special points of a quadratic- There are 3 key points of a quadratic that may have special meaning:1) vertex 2) y-intercept 3) x-intercept

In the flight of a golf ball there are three questions that are often asked:1) What is the highest point in its flight? This can be read from the graph and is the y-coordinate of the vertex which is the maximum value of the graph.2) How far off the ground was the football when it was thrown? This corresponds to the y value of the y-intercept.3) After how many seconds will the ball hit the ground? This corresponds to the x value of the x-intercept.

The graph below illustrates these key points and what they represent.

From the graph above:1) The maximum height of the football is 51 meters (y value in vertex)2) The height above ground when hit is 9 meters (y value of y-intercept)3) The time it takes to hit the ground is 6.2 seconds (x value of x-intercept)

Vertex (2.9, 51)Maximum value = 51

y intercept (0,9) Height above ground is 9 m

x intercept (6.2 , 0)time to hit groundis 6.2 seconds


Using the graphing calculator these values may be obtained using the following keystrokes:1) 2nd trace 4: maximum (maximum height of football)2) 2nd trace 1: value (height above ground when thrown)3) 2nd trace 2: zero (time it takes to hit the ground)

Asn(A9) Handout of 2 questions with table of values that require Quadratic Regression Analysis, maximum value, y intercept and x intercept determination. (see end of this document)

Test #1

Asn(A10) handout on campfire sheet.Asn(A11) Q 24,25,26 p. 22 Advanced Q 23 a) b) p. 22Asn(A12) Handout of problems

Math 112 Quadratics- NotesAsn(B1) Transformations handoutAsn(B2) Practice of transformations (to be submitted for marking)

Note(22) Forms of quadratic functions- Quadratic functions are found written in 3 main forms:

1. transformational form: 1 (y - k) = ( x - h)2 a 0 e.g. 1 (y + 2) =(x - 3)2

a 3

2. standard form: y = a(x - h)2 + k a 0 e.g. y = 3(x - 3)2 - 2

3. general form: y = ax2 + bx + c a 0 e.g. y = 3x2 - 18x + 25

Each form provides useful immediate information about the graph such as:1) the location of the vertex2) the location of the y-intercept3) vertical stretch of the graph4) vertical and horizontal translations of the graph5) maximum/minimum

It is useful to learn each form of graph as each reveals its own characteristics of the quadratic function. Each may be rewritten in the other form by algebraic method.

Note(23) The transformational form- This form of the function has the general formula:

1 (y - k) = ( x - h)2 a 0a

The meaning of the letters a, k., and h are:a= vertical stretch of the graph. (take the reciprocal of 1/a) k= the vertical translation of the graph (multiply by -1)h= the horizontal translation of the graph (multiply by -1)


Note(24) The standard form- this form of the quadratic function has the general formula:

y= a(x-h)2 + k where a 0

The meaning of the letters a, h, and k are the same:

a = vertical stretch factor. (read as is)h = horizontal translation (multiply by -1)k = vertical translation (read as is)

The advantage of the standard form of the function is:1) It gives the vertex of the function…. (h, k)2) It gives the vertical stretch factor as……. a3) It can be easily be entered into the graphing calculator

Asn(B3) Writing a quadratic in standard form. (see end of this document)


Note(25) The general form- This form of the quadratic function has the formula:

y = ax2 + bx + c a0

The meaning of the letters a and c are:a = vertical stretch factorb = (unknown)c = y- intercept value

The advantage of the general form of the function is:1) It gives the vertical stretch factor as…….. a2) It gives the y-intercept of the graph as…...c3) It is the form given by the calculator and may easily be entered.

½ (y+1) = (x +2)2 transformation formaty = 2(x+2)2 -1 standard format

To write in general format the square (x+2)2 must be expanded:

y = 2(x+2)2 -1 expand (x+2)2 y = 2[(x+2)(x+2)] - 1 use FOIL to expandy = 2[x2 +2x +2x + 4] - 1 add middle two “x” termsy = 2[x2 +4x + 4] - 1 distribute the 2 in fronty = 2x2 + 8x + 8 - 1 simplify the constants

y = 2x2 + 8x + 7 a = 2 the vertical stretch

Asn(B4) Writing a quadratic in general form. (see end of this document)

c = +7 the y-intercept (0,7)

Math 112 Quadratics- NotesNote(26) In order to rewrite the general form of an function into transformation form it is necessary to recognize a “perfect square trinomial”.

To make a perfect square trinomial a binomial is squared.e.g.(x +3)2 = x2 + 3x + 3x +9

= x2 + 6x +9 Perfect square trinomial

A perfect square trinomial can be factored into two identical factors:e.g. x2 + 6x +9 = (x +3) (x +3)

= (x+3)2

We recognize (x+3)2 as the factored form of the “perfect square trinomial” and as the “x- part” of the transformational form of a some quadratic function.e.g ½ (y +3) = (x+3)2

From the above information we conclude that the general form of an function must be written in such a way as to contain a perfect square trinomial which is then factored into a form similar to (x+3)2 above. It is necessary, therefore, to create a perfect square trinomial as the first step in rewriting.

From this example “ x2 + 6x +9 “ we compare the 6 with 9 and notice that if we take ½ of 6 then square the answer we obtain 9. I.e. (6/2)2 = 9. From this we have a way of predicting the third term of any perfect square trinomial.

In general “c” in ax2 + bx + c can be found as c = (b/2)2 .Eg. Find “c” in x2 + 10x + c.c = (b/2)2

c = (10/2)2

c = (5)2

c = 25so that x2 + 10x + 25 is a perfect square trinomial with factors:(x+5)(x+5)(x+5)2

Asn(B5) Q 4 p. 29 jumbled equationsAsn(B6) Q 3 p. 29 reflection on x- axisAsn(B7) Extra practice. (summary handout)


Note(27) Writing the general form into transformational form requires that we find the value of “c” for a quadratic. This is called “completing the square”.

e.g Rewrite this function in transformational form by completing the square.

y = x2 + 6x + 12 Move the constant to left hand side bysubtracting 12 from both sides.

y - 12 = x2 + 6x Find the value of c to complete thesquare. c = (b/2)2 so (6/2)2 so c = 9

y - 12 + 9 = x2 + 6x + 9 Factor the trinomial, and simplifyconstants on left side.

(y-3) = (x+3)2 Transformational form.

Some quadratics written in general form require GCF factoring first.e.g. Rewrite this function in transformational form by completing the square.

y = 2x2 -8x + 5 Move constant to left side of the bysubtracting 5 from both sides.

y -5 = 2x2 - 8x Factor out 2 on right hand side.

y - 5 = 2[x2 - 4x] Find the value of c to complete thesquare. c = (b/2)2 so (-4/2)2 so c = 4

y - 5 +2 4 = 2[x2 - 4x + 4 ] Add 2 times 4 (equivalent to 8) toboth sides of the equation.

y - 5 + 8 = 2[(x -2)2 ] Complete the square and simplifyconstants on the left side.

1 (y +3) = 1 2(x - 2)2 Multiply both sides by the reciprocal 2 2 of 2 or ½ ½ (y +3) = (x -2)2 Transformational form Asn(B8) Q 15, 17, 20 p. 33 Advanced: Q 32 p. 35

Asn(B9) Q 19, 22, 26 a) and c) p. 33, 34

Asn(B10) Handout on completing the square. Advanced Q31 p. 35


Note(27) (repeated). Writing the general form into transformational form requires that we find the value of “c” for a quadratic. This is called “completing the square”.

An alternative method of completing the square is provided here for the same example given on the previous page.

y = 2x2 -8x + 5 Divide all terms by “a” the first termcoefficient. “a” =2 in this case

y = 2x2 -8x + 5 2 2 2 2y = 2x2 -8x + 5 Reduce all terms if possible2 2 2 21 y = x2 -4x + 5 Simplify2 21 y - 5 = x2 -4x + 5 - 5 Subtract 5/2 from both sides2 2 2 2

1 y - 5 = x2 -4x + 5 - 5 Simplify right side.2 2 2 2

1 y - 5 + 4 = x2 -4x + 4 . Make a perfect square trinomial using2 2 (b/2)2 = (-4/2)2 = (-2)2 = (-2 -2) = +4

1 y - 5 + 4 = (x -2)(x -2) . Factor the right side.2 21 y - 5 + 4 = (x -2)2 . Write as a single binomial squared.2 21 y - 5 + 8 = (x -2)2 . Find a common denominator2 2 2 for -5/2 and 4. (Rewrite 4 as 8/2).1 y + 3 = (x -2)2 Write left side in brackets.

2 21 (y + 2 3 ) = (x -2)2 Use the reciprocal of ½ (outside the2 1 2 bracket) to determine the value

needed inside 1 (y +3) = (x -2)2 Transformational form


Note(28) Determining the equation of a quadratic from a graph.

Method #1:1) Find the coordinates of the vertex (h , k)2) Place k and h into the transformation formula3) Solve for the unknown stretch factor in front (?) of the transformation formula by substituting in a second point from the graph.

e.g. Find the equation of the quadratic illustrated below:

transformation formula:1/a(y - k) = (x - h)2 Substitute in vertex (3,4)(?)(y - 4) = (x - 3)2 Replace 1/a with (?) for simpler solving (?)(y - 4) = (x - 3)2Substitute in another point (4,2) (?)(2 - 4) = (4 - 3)2

(?)(-2) = (+1)2 Solve for (?) (-2)(?) = 1

(?) = - 1 2

Replace (?) with -1 to obtain formula and use vertex (3,4) 2

-1 (y - 4) = (x - 3)2 Function in transformational format 2

Asn(B11) Q 39, 40 p. 38 39

(3,4)

(4,2)


Note(29) Determining the equation of a quadratic from a graph.

Method #2:1) Find the coordinates of the vertex (k , h)2) Locate another point through which the graph passes.3) Determine the stretch factor “a” by comparing to an equivalent point of the basic quadratic. (Divide equivalent y coordinates.)4) Determine the sign of the quadratic by the direction the graph opens.5) Enter k, h, a, and the sign into the transformational formula.

e.g. Find the equation of the quadratic illustrated below:

transformation formula:1/a(y - k) = (x - h)2 Substitute in vertex (3,4)1/a(y - 4) = (x - 3)2 Determine the stretch factor using point

(4,2):For y = x2 (over 1, up 1)For this graph (over 1, down 2)Stretch factor: 2 divided by 1 = 2

1 (y -4) = (x - 3)2 Determine the sign (+) or (-) in front2 of 1/a. Since the graph turns down

the sign is negative. -1 (y - 4) = (x - 3)2 Transformational format 2Asn(B12) Q 1 p. 61 in curriculum document.Test #2

(3,4)

(4,2)


Note(30) Quadratic functions come in the form:y = ax2 + bx + c where a 0

Very often a quadratic function will touch or cross the x- axis. The point(s) where the function crosses are significant as they often have special meaning. These points contain the zeroes or roots of the quadratic.

y = x2 +1 no x interceptsno zeroes of the quadratic functionno Real roots, the roots are imaginary

y = x2 one x interceptone zero of the quadratic functionone root of the quadratic equation


y = x2 -3 two x interceptstwo zeroes of the quadratic functiontwo roots of the quadratic equation

Roots or zeroes

For the above graph the points on the x-axis are called intercepts.The x values of these points are called:

1) the roots of the quadratic equation or2) the zeroes of the quadratic function


Note(31) Quadratic Equation- a quadratic equation comes in the form:0 = ax2 + bx + c where a 0 and y= 0

When the quadratic function has y set to 0 it is called a quadratic equation. Solving for the variable “x” gives us the values of the “roots” of the equation or “zeroes” of the function.

For the quadratic function y = x2 +2x –8, the roots can be found by solving the quadratic equation when y is set to “0”. We solve for variable in the equation: 0 = x2 +2x –8 and obtain the roots: x = 2 and x = -4. These values may be obtained a number of ways….

Method #1 Graphing calculator methodEnter the function after y1= x2 + 2x –8 Press 2nd trace option 2:zeroes. Set left boundary, enter, and right boundary, enter, then press enter to guess the value of the root.

Method #2 Factoring methodSet y = 0 then factor the quadratic equation0 = x2 +2x –8 0 = (x-2) (x+4) set x-2 = 0 set x+4 = 0

x = 2 x = -4The roots of the equation are 2 and –4The submarine will surface in 2 hours. (-4 is an inadmissible time)

Method #3 completing the square method

0 = x2 +2x –8 Add 8 to both sides to move constant8 = x2 + 2x Find the missing 3rd term of the trinomial8 + 1 = x2 +2x + 1 (b/2)2 = (2/2)2 = (1)2 = 19 = (x+1)2 Take the square root of both sides3 = (x+1) Subtract 1 from both sides

-1 3 = xThe roots are: x = -1 +3 and x = -1 – 3

x = +2 x = -4The submarine will surface in 2 hours. (-4 is an inadmissible time)

Asn(C1) Q 8 a c d ,9 a b ,10 a) g) p. 45


Method #4 using the quadratic formula

The quadratic formula obtained from completing the square using the general form of the quadratic y = ax2 +bx +c is:

x = -b (b) 2 – 4ac 2a

For the equation y = x2 +2x –8 the values of a, b, and c are a =1 b = 2 c= -8Substituting into the formula we have:x = - (2 ) (2 ) 2 – 4( 1 )(-8 ) x = -2 + 6 = 4 = +2

2( 1 ) 2 2x = -2 4+32

2 x = -2 – 6 = -8 = -4x = -2 36 x = -2 6 2 2

2 2The roots are +2 and –4

The submarine will surface in 2 hours. (-4 is an inadmissible time)

Asn(C2) Q 24 p. 48 Advanced: Derive the quadratic formula by hand


Note(32) Using the quadratic formula to find the two roots of the equation gives us the x values of the intercepts. Averaging these values will give us the center point of the parabola along the x- axis through which the axis of symmetry runs. The vertex shares this same x-value. The average for the two x values expressed in general form is:

-b + (b) 2 – 4ac + -b - (b) 2 – 4ac 2a 2a = -b

2a 2

This expression (–b/2a) is the midpoint of the parabola along the x axis and therefore the x value in the vertex. We may find the vertex by solving for x then substituting this value into the quadratic function to find the y value. e.g. Find the vertex of the parabola y = x2 +2x – 8.

Midpoint = -b = -(2) = -2 = -1 Coordinate of midpoint (-1, 0) 2a 2(1) 2Since the vertex is directly below/above the midpoint, the coordinate of the vertex is: (-1 , ?)Now substitute –1 into the quadratic function to find the y value of the vertex.

y = x2 +2x -8= (-1 )2 + 2(-1 ) -8= 1 – 2 – 8

y = -9

The vertex has coordinates (-1,9)

Asn(C2) cont’d Q 25 p. 48


Note(33) The quadratic formula will sometimes produce an answer that does not belong to the Real number system. In such a case the quadratic has no Real roots so that the graph has no x-intercepts. The roots are complex numbers and belong to the imaginary number system.

E.g. Find the roots of the equation 0 = x2 –6x +13

Use the quadratic formula:x = -b (b) 2 – 4ac

2ax = - (-6 ) (-6 ) 2 – 4(1 )(+13 ) a= 1 b = -6 c = +13

2(1 )x = +6 36 – 52 2x = +6 -16 Rewrite -16 as 16 -1 2x = +6 16 -1 2x = +6 16 -1 Separate 16 and –1 into two radicals 2x = +6 4 -1 Take the square root of 16 2x = +6 4i Write -1 as i, an imaginary number 2 x = +3 2i Reduce by dividing each term by 2

The roots of the quadratic are:X = +3 + 2i and x = +3 – 2i

Note: These are imaginary roots and therefore there are no x-intercepts. Asn(C3) Q 27,28,29 p. 48,49

Asn(C4) Q 30 a,c,e,g 31 a,b p. 49 Advanced Q 32 a , c, e p. 49

Math 112 Quadratics- NotesNote(34) Problem solving often requires that we solve for the vertex and/or the roots of the equation. The vertex gives the maximum or minimum value of a question whereas the roots give the “time to hit the ground” or the value when there is “no profit” or “width” when the area is zero etc.

Eg. a golf ball is hit from a 1 meter platform inside a golf dome and follows a path given by y = -6x2 +12x +1 where x = time (s) and y = height (m)

1) What maximum height will it reach? (Hint: find the vertex)Vertex formula = -b = - (12) = -12 = +1 = vertex = (+1 , ?) 2a 2(-6) -12To find the y value in the vertex we substitute +1 into the formula as follows:

y = -6x2 +12x + 1y = -6(+1)2 +12(+1) +1y = -6(+1) + 12(+1) +1y = -6 +12 +1y = +7The vertex has coordinates (1,7) which we interpret as after 1 second

the golf ball reaches a maximum of 7 meters.

2) When will it hit the ground? (Hint: It hits the ground at the x- intercept)Find the zeroes or roots of the quadratic equation:

x = -b (b) 2 – 4ac 2a -12 13 -12 +13 = 1 = -0.8

x = - (12 ) (12 ) 2 – 4(-6 )(1) -12 -12 -12 2(-6) x = -(12) 144 +24) -12 –13 = -25 = +2.08 -12 -12 -12x = -12 168 -12

The golf ball will hit the ground after 2.08 seconds. The first answer is inadmissible as it is not sensible to have a negative answer.Asn(C5) Q 22 p. 33, Q 26 p. 34, Q 5 p. 44, Q 39,40,41 p. 53Asn(C6) Q 34,35,36,38 p. 52,53

Math 112 Quadratics- NotesNote(35) The discriminant of the quadratic formula is given by the expression b2 – 4ac. The value of the expression can assist us in determining the nature of the roots for a quadratic equation.

e.g Find the discriminant for the following quadratic by using the expression b2 –4ac . Also find the roots of the equation.Example #1 y = x2 –6x +13 a = 1 b = -6 c = 13

The value of the discriminant is:b2 –4ac(-6 )2 – 4(1 )(13)36 – 52-16 Note: The discriminant is negative

Find the roots:x = -b (b) 2 – 4ac 2ax = -(-6 ) -16 (from above) 2(1)x = +6 16 -1 -16 may be written as 16 -1 2x = +6 16 -1 16 -1 may be separated into 16 -1 2x = +6 4i Take the square root of 16 and replace -1 with i 2x = +6 4i Separate into two fractions and reduce 2 2x = 3 2i which gives the 2 roots…… 3 + 2i and 3 – 2iConclusion: When the discriminant is negative we obtain 2 complex roots. The graph obtained indicates that there are no x-intercepts:

Math 112 Quadratics- NotesExample #2 y = x2 +4x + 4 a = 1 b = +4 c = 4

The value of the discriminant is:b2 –4ac(+4 )2 – 4(1 )(4)16 160 Note: The discriminant is zero

Find the roots:x = -b (b) 2 – 4ac 2ax = - (4 ) 0 (from above) 2(1)x = -4 2x = -2 One answer

By factoring we discover that there are two roots (not one) but they are identical. We can factor the trinomial to see the two roots:

0 = x2 +4x +40 = (x +2)(x + 2)

First factor: Second factor:We set x +2 = 0 We set x +2 = 0Solve: x = -2 Solve: x = -2

Conclusion: When the discriminant is zero we obtain 2 real roots that are identical. The graph obtained indicates that there is one x-intercept:

(-2 , 0)


Example #3 y = x2 -x -6 a = 1 b = -1 c = -6

The value of the discriminant is:b2 –4ac(-1)2 – 4(1 )(-6)1 +2425 Note: The discriminant is positive

Find the roots:x = -b (b) 2 – 4ac 2ax = -(-1 ) 25 (from above) 2(1)x = +1 5 2x = 6 = +3 or x = -4 = -2 The roots are +3 and -2 2 2

Conclusion: When the discriminant is positive we obtain 2 real roots. The graph obtained indicates that there are two x-intercepts.

(0,-2) (0,+3)

Summary of the discriminant:

If b2 – 4ac < 0 The graph has two imaginary roots and no x-intercepts.If b2 – 4ac = 0 The graph has two identical real roots and one x-intercept.If b2 – 4ac > 0 The graph has two real roots and two x-intercepts.

Note: All quadratics have two roots. It is the x-intercepts that vary.

Asn(C7) Q 53,56,57, 58 a b, 59 p.56,57Asn(C8) Handout on various problems involving quadratics. Selected.

Test #3

Asn(C9) Advanced Q 62,64,63 p. 57 Advanced Q Investigation #7 65,66,67 p. 58

Advanced Q 68 to 72 p. 59 selected questions.

Assignment section for handouts

Asn(A7) Extra practice sheet (to be submitted for marking)

1. Identify the type of sequence as arithmetic, quadratic, cubic or none of these:

A) [-2, -5, -8, -14, …]B) [-3, -10 , -21, -36, -55, -78, -105, …]C) [–1, -14, -51, -124, -245, -426, -679,…]D) [-1/3, 2/3, 3, 20/3, 35/3, 18, 77/3, 104/3, 45, …]E) [11, 101, 1001, 10001, 100001, …]F) [1/5, 2/5, 4/5, 7/5, 11/5, 16/5, …]G) [0.0, 3.2, 18.4, 75.6, 252.8, 700.0, 1663.2, 3508.4, 6745.6, …]

2. Find the first four terms of each sequence:

A) tn = -2n – 5B) tn = -3n + ¾C) tn = -n2 –5n +3D) tn = 1/3 n2

3. What is D1 for the sequence defined by tn = 5/7n +4/5

4. Dan had 80 m of fencing to enclose a rectangular yard on three sides. He used his house as the fourth side. By experimenting with different widths as shown in the table, Dan found he could enclose different areas. What type of function could he use to model the data/

Width 1 2 3 4 5 6 7 8Area 78 152 222 288 350 408 462 512

Asn(A9) Determining the special points of a quadratic

1. The table shows the height of a golf ball (in metres) after various lengths of time measured in seconds. The tee-off is on a cliff.

Time (s) 1 2 3 4 5Height (m) 32.8 46.8 51.0 45.4 30.0

a) Draw a sketch of your graphb) Does the data suggest a linear or quadratic function? Explainc) What is the maximum height of the golf ball in its flight?d) What is the height of the cliff?e) How long will it take to hit the fairway?f) State the Domain and Range of the graph

2. The table below gives the height of a baseball (in metres) for various times measured in seconds.

Time (s) 1 2 3 4 5 6Height (m) 32.3 53.3 64.5 65.9 57.5 39.3

a) Draw a sketch of your graph.b) What is the value of D2?c) What is the maximum height of the baseball in its flight?d) How long does it take to reach this height?e) How far off the ground is the baseball when it is hit?f) After how many second will it hit the ground?g) State the Domain and Range of the graph

Asn(B3) Write the following in standard form and interpret the standard form by identifying the vertex and vertical stretch factor.

1. ½ (y-1) = (x – 1)2

2. 2(y – 1) = (x + 2)2

3. ¼(y = (x+2)2

4. 3(y-2) = (x)2

5. 2y = x2

Asn(B4) Write in general form and identify the y- intercept and vertical stretch factor.

1. ½(y-1) = (x –1)2

2. 2(y+1) = (x+3)2

3. ¼(y) = (x+2)2

4. 3(y+2) = x2

5. 2y = x2

Math 112 Rate of Change- Notes

Note(36) Rate of change is a concept that is used in many different contexts. A few examples are given below:

1. Birth rate births/1000 per year2. Heart rate beats/ minute3. Population growth rate people/year4. Employment rate people/year5. Global warming degrees Celsius/ year6. Rate of travel (speed) km/hour7. Production rates units/hour

Each of these rates can be calculated by dividing the first quantity by the second which is given as a time unit.

Eg. #1 A car travels 220km in 2 hours. What is the rate of travel in km/hour?

Rate = D/t = 220/2 = 110km/hr

E.g. #2 Your heart beats 20 times in 15 seconds. What is your heart rate in beats/min?

Heart rate = 20 beats/15 sec = 1.3 beats/sec = 80 beats/min.

Asn(D1) Handout


Note(37) Function notation-

The general notation of y = ax2 + bx + c is often seen written in function notation as: f(x) = ax2 + bx + c

Since y = f(x) we may write the general form of a point (x,y) as ….. (x, f(x) ) so that (x,y) = (x , f(x) )

For the 2 general points (x1,y1) and (x2, y2) the slope is given by:

Slope = Δ y = y2 - y1 Δ x x2 – x1

The same 2 points written in function notation are:

(x1 , f(x1) ) and (x2 , f(x2) ) and have the slope given by: Slope = Δ f(x) = f(x2)- f(x1) Δ x x2 – x1

For points (a, f(a)) and (b, f(b)) the average rate of change is often seen written as:

Δ f(x) = f(b) – f(a)Δ x b - a


Note(38) The line drawn between two points on a curve is known as the secant for the two points. The slope of the secant is equal to the average rate of change.

B A

Secant through points A and B

The average rate of change for a function can be calculated by using the formula:

Δ f(x) = f(x2)- f(x1) Δ x x2 – x1

E.g. John leaves for work to go to Toronto at 6 am and arrives at 8 am . The odometer is reset to read 0 km at the start. At the end of his trip it reads 240 km. What is his average rate of travel?

Points are (6, 0) and (8,240) where x = time and y = distance

Average rate = Δ f(x) = f(x2)- f(x1) = 240 – 0 = 240 = 120 km/hr of change Δ x x2 – x1 8 –6 2

Asn(D2) Handout

Asn(D3) Q 13 a,b p. 79 Q 15 a,b,c p. 80 Q 16 a to e p. 81 Q 17 a,b,c p.81 Q 24 a to c p. 83

Asn(D4) Handout. Curriculum Document


Note(39) The average rate of change for a straight line is a constant regardless of points used to calculated.

e.g The function f(x) = 2x + 1 has ordered pairs such as:A (0,1) B(1,3) C(2,5)The graph would appear as follows:

C Ave rateAB = 3 – 1 = 2 = +2 B 1 – 0 1

A Ave rateBC = 5 –3 = 2 = +2

2 – 1 1

The average rates are both +2 thereforeconstant. (A straight line has only one slope)

Note(40) The average rate of change for a quadratic is not a constant but is continually changing.

e.g. The path of a football is given by the function f(x) = -1/2 x2 + 4x -2 and has ordered pairs such as:A (1, 1.5) B(2,4) C(3, 5.5) D (4,6) E(5, 5.5) F(6,4)G(7,1.5)

Ave rateAB = 4.0 -1.5 = +2.5 2 - 1Ave rateBC = 5.5 –4.0 = +1.5 Ave rate at vertex = 0 3 - 2Ave rateCD = 6.0 –5.5 = +0.5 4 – 3 D EAve rateDE =5.5 –6.0 = -0.5 C F 5 – 4 B GAve rateEF = 4.0 - 5.5 = -1.5 6 – 5 A FAve rateFG = 1.5 - 4.0 = -2.5 7 - 6


Note(41) The average rate of change for a football’s flight is:

1) Positive as it gains height2) Zero at the maximum height3) Negative as it loses height

Note(42) Instantaneous rate of change is a concept that asks for the rate of change for a single point in time. It is not an average.

An example of where this can be illustrated in real life is asking for the speed at any given time when a car is accelerating. By looking at your speedometer we can read the instantaneous rate of speed at any time. The speedometer measures instantaneous rate of change.

Asn(D5) Handout for calculating the average rate of change as P approaches Q where the function is y = x2 and the point of interest is (5,25)

Note(43) We can approximate the instantaneous rate of change for a function by using 2 points either side of a given point of interest that are very, very close to the given point

E.g. Find the instantaneous rate of change for the function y =x2 when x = 5

Step #1 Find two points on either side of (5,25) the point of interest that fit the curve y= x2

If we choose the increment (.01) we can obtain points close to (5,25) by:

Adding .01 to 5 to obtain the point (5.01, ? )Subtracting .01 from 5 to obtain the point (4.99, ? )

We may now find the y value of each point above by substitution:

y = x2 so that (5.01)2 = 25.1001 which gives us (5.01, 25.1001)y = x2 so that (4.99)2 = 24.9001 which gives us (4.99, 24.9001)


The two points are equidistant from the point of interest (5,25) and if connected with a straight line form a secant line that is very close to the tangent at the point (5,25). This line is parallel to the tangent at this point. Since parallel lines have equal slopes the slope of the secant would equal the slope of the tangent at point (5,25).

Step #2 Using the formula for the slope of a function we obtain an approximation for the instantaneous rate of change at point (5,25)

Point (5.01, 25.1001)Point (4.99, 24.9001)

Instantaneous rate = f(b) – f(a) = 25.1001 – 24.9001 = .2 = 10of change b a 5.01 – 4.99 .02

The instantaneous rate of change for the function y =x2 at the point (5,25) is 10

Asn(D6) Approximating instantaneous rate of change. (overhead).Asn(D7) Questions on instantaneous rate of change. (handout NFLD)Asn(D8) Garden Question (handout).Asn(D9) Curriculum Document (handout).


Note(44) Problem solving may require that you find one of the following points of interest:

Point of interest: When you find these points:

a) vertex If the question asks for a maximum (height, area) b) x-intercepts If the question asks for time in airc) y-intercepts If the question asks for initial starting point

Point of interest: Formulas to use:

a) vertex (-b , ) Once you find the x coordinate substitute 2a into the quadratic to obtain the y- coordinate

b) x-intercepts x = -b b 2 –4ac 2a

c) y-intercepts Read this from the quadratic. It is the valuec in the function if it is written in generalformat y= ax2 + bx + c

Example:

Babe Ruth was renowned for hitting high fly balls in baseball. One day he made contact with a ball 1 metre above ground and hit the ball at a speed of 27.5 m/sec. Using the general formula h = H + vit – 4.9t2 substitute for H and vi to make a quadratic function to fit this situation.

a) What maximum height did the ball reach?b) When did it reach this height?c) How high off the ground was the ball hit?d) How long was the ball in the air?e) What was the ball’s average velocity between 1 and 2 seconds?f) What is the height of the ball at 4 seconds?g) What was the speed of the ball at 4 seconds?


To solve this problem we need to substitute for the values H and vi H = initial height = 1 metre (we are careful not to confuse H for h)Vi = initial velocity = 27.5 m/sec

By substitution we have:h = 1 + 27.5t –4.9t2

Written in descending order of power we have:h = -4.9t2 + 27.5t + 1

For the purposes of the quadratic formula and vertex formula we identify the letters a, b, and c as:

a = -4.9b = 27.5c = 1

We are now ready to answer the question a) through f)

a) What maximum height did the ball reach?

This question requires that you solve for the vertex. The y value of the vertex gives the maximum height. Use the formula: –b = -27.5 = -27.9 = + 2.846938776 2a 2(-4.9) -9.8

We are careful to avoid rounding off too early so choose 2.85 as the x coordinate. Now we substitute this value into the quadratic h = -4.9t2 + 27.5t + 1

h = -4.9(2.85)2 + 27.5(2.85) + 1h = 39.57 m

The vertex is therefore (2.85, 39.57). The y-coordinate is 39.57m, the maximum height of the ball.

Statement: The maximum height is 39.75 metres


b) When did it reach this height?

This question requires that you have the vertex of your quadratic which you have already found in a) above (2.85, 39.57). The x coordinate is the answer to this question 2.85

Statement: The time required to reach this maximum height is 2.85 seconds.

c) How high off the ground was the ball hit?

This question does not require a formula. The c value in the quadratic gives this answer, as it is the y-intercept value.

h = -4.9t2 + 27.5t + 1 The value of c is +1.

Statement: The ball was hit 1 metre off the ground.

d) How long was the ball in the air?

The question is asking for the time, which is given along the x-axis. Therefore, we need the distance along the x-axis which is given by the x coordinate of the x intercept. We use the quadratic formula.

x = -b b 2 –4ac 2a x = -55.35 = 5.64x = -27.5 (27.5) 2 –4(-4.9)(1 ) -9.8 2(-4.9)x = -27.5 775.85 -9.8x = -27.5 27.85 x = +.35 = -0.0357 -9.8 -9.8

The value 5.64 is the x value of the x-intercept (5.64,0) and represents the time of the ball in the air.

Statement: The ball was in the air for 5.64 seconds.


e) What was the ball’s average velocity between 1 and 2 seconds ?

The word average suggests two items or points are involved. This requires that we find or locate two points for which an average velocity is to be calculated from. The numbers 1 and 2 are in fact the x-values of the two points… (1, ) and (2, ).

Two find the y values of these points we substitute into the quadratic as follows:

x = 1 or ( t = 1) x = 2 or (t = 2)

h = -4.9t2 + 27.5t + 1 h = -4.9t2 + 27.5t + 1h = -4.9(1)2 + 27.5(1) + 1 h = -4.9(2)2 + 27.5(2) + 1h = 23.1 h = 35.4

Point (1, 23.1) Point (2, 35.4)

The average rate of change is found using the equivalent of the slope formula but for quadratic functions;

f(b) – f(a) = f(2) – f(1) = 35.4 – 23.1 = 12.3 = 12.3 m/sec b- a 2 – 1 2 – 1 1

Statement: The average rate of change for t = 1 to 2 seconds is 12.3 m/sec.

Math 112 Rate of Change- Notesf) What is the height of the ball at 4 seconds?

This question requires that we understand that 4 represents the time and that we are asking for height so that we substitute 4 for t and solve for h.

Substitute t = 4 into the quadratic to obtain the y valueh = -4.9t2 + 27.5t + 1h = -4.9(4 )2 +27.5(4 ) + 1h = 32.6Point (4, 32.6)Statement: The height of the ball at 4 seconds is 32.6 metres.

g) What was the speed of the ball at 4 seconds?This question is in fact asking for the instantaneous rate of change

when t = 4 seconds and must be distinguished from the question “What is the height of the ball at 4 seconds?” The key word here is speed

Step #1 Find two points really close to (4,32.6) by adding and subtracting a small increment (.01) to/from the x value of this point.4 + .01 = 4.014 - .01 = 3.99 The two points are: (4.01, ) and (3.99, ). Now substitute to find the y-values of each point. (Do not round off the y-values for accuracy’s sake).

Point (4.01, ) Point (3.99, )h = -4.9t2 + 27.5t + 1 h = -4.9t2 + 27.5t + 1h = -4.9(4.01)2 +27.5(4.01) + 1 h = -4.9(3.99)2 +27.5(3.99) + 1h = 32.48251 h = 32.71651

The points are: (4.01, 32.48251) and (3.99, 32.71651)Step #2 Find instantaneous rate of change or speed use the formula f(b) – f(a) = 32.71651 – 32.48251 = 0.231 = 11.55 m/s b - a 3.99 – 4.01 0.02Statement: The instantaneous rate of change (speed) at the time t = 4 seconds is 11.55 m/s.

Test #4

Math 112 Exponential GrowthNote(45) A geometric sequence is a sequence in which each term can be generated by multiplying each successive term by a common factor. e.g 2 , 4 , 8 , 16, This geometric sequence is generated by multiplying 2 2 2 each term by the common factor 2 Note(46) A common ratio is a factor used to generate a geometric sequence.The common ratio above is 2.

Note(47) We can find the common ratio of a sequence by dividing any two successive terms as long as we use the current term as the dividend and the previous term as the divisor.

e.g Find the common ratio for the following sequence:3, 12, 48 ,192

12 = 4second term divided by first term 348 = 4third term divided by second term12

Note(48) A recursive formula is a formula that is made by relating two successive terms in some way such that other successive terms are related in the same way.

To develop a recursive formula we first establish the pattern between any two successive terms in a concrete way then generalize using variables.

E.g. Develop a recursive formula for this geometric sequence:2 , 6, 18, 54The common ratio is 6/2 = 3 so that b = 3 therefore:2nd = three times the 1st 6 = (3) 2 Replace 2nd term by 6 and 1st by 218 = (3) 6 Replace 3rd term by 18 and 2nd by 6tn+1 = (3) tn Replace next term by tn+1 and current

term by tn

The recursive formula for this sequence is tn+1 = (3) tn

Math 112 Exponential Growth

Note(49) A general formula for a geometric sequence is much more practical that the recursive formula as the general formula will give the value for any term without the need for a previous term. The general formula for a geometric sequence is given by y =abx

where:a = the first termb = the common ratiox = real number

e.g Generate a geometric sequence where the first term is 3 and the common ratio is 4. a = 3 the first termb = 4 the common ratio

Substituting the values of “a” and “b” into the formula we have:y = a*bx

y = 3*4x

x value substitution y- value

Let x = 0 y = 3*4x = 3*40 = 3*1 = 3 the 1st numberLet x = 1 y = 3*4x = 3*41 = 3*4 = 12 the 2nd numberLet x = 2 y = 3*4x = 3*42 = 3*16 = 48 the 3rd number

The sequence generated is:

3 , 12, 48, etc.


Note(50) We can determine the general formula for a geometric sequence… y = a*bx by identifying the 1st term “a” and by calculating the common ratio.Eg. What is the general formula that will generate the sequence below:2, 6, 18, 54, 162, etc.a = 2 the first termb = 6/2 = 3 the common ratio

Substituting the values for “a” and “b” into the general formula:y = a*bx

y = 2*3x The general formula for the sequence is y = 2*3x

Note(51) Graphing technology can generate the same sequence and ordered pairs can be read from the resulting table.

Enter the equation after \y1= 2*3x Press 2nd Table to see the ordered pairsPress Graph to see the graph of the function

Asn(E1) Investigation #1 Q 3 and 8 p. 110 to 113


Note(52) The graph of a geometric sequence is discrete with a domain given by D = {x x N}

E.g Graph the following:

X 1 2 3 4Y 2 4 8 16

Discrete = dotted line graphing 6 5 4 3 2 1 1 2 3

a) What is the recursive formula for this sequence? By observing the pattern in the second row we notice that the next term is 3 times the previous. By example we have:2 = (2) 1 the 2nd is double the 1st

4 = (2) 2 the 3rd is double the 2ndtn+1 = (2) tn the next is double the previous

The recursive formula is : tn+1 = (2) tn

b) What is the y intercept? The first term of the sequence has x = 0 and y = 1

The y intercept is (0,1) y = 1

c) What is the common ratio? The common ratio is found by dividing any two successive terms. So in general the common ration is found by:tn+1 tn = 2/1 or 4/2 = 2

d) What is the range of the graph? R = {y y 2, y N}Math 112 Exponential Growth

Note(53) The graph of a geometric function is continuous with a domain given by D = {x x R}

E.G. Graph y = 3x

Continuous = solid line graphing

a) What is the general formula for this graph? y = 3x

b) What is the y intercept? Given when x = 0 so y = 30 = 1The y intercept is y = 1

c) What is the common ratio? The base is the common ratio so = 3

d) What is the Range for this graph? R = {y y>0, yR}


Note(54) Growth and decay curves are determined by the “b” value in the general formula y = a*bx

For the general formula y = a*bx

a) if b < 1 the curve is a decay curve. E.g. let b = (0.5)e.g. y = 2* (0.5)x

b) If b>1 the curve is a growth curve. E.g. let b = (1.5)e.g. y = 2(1.5)x

Asn(E2) Investigation #2 Q 17 to 21 p. 115


Note(55) Function notation for geometric sequences can be written n the form:

f(x) = a*bx where a = first term b = common ratio x 0

E.g. A geometric sequence is given by : 2, 6, 18, 54 a) Write this in function notationb) Find f(3) c) What is the value of f(5)?

a) a = 2 the first term. b = 6/2 = 3 the common ratioThe function notation for this geometric sequence is f(x) = 2*3x

b) f(3) = 2*3x Substitute in x = 3 f(3) = 2*33

f(3) = 2*27 f(3) = 54

c) f(5) = 2*3x Substitute in x = 5 f(5) = 2*35

f(5) = 2*243 f(5) = 486

Exponential Growth

Note(56) Geometric sequences relate to many real life situations such as:

1) Population growth2) Radioactive decay3) Compound interest growth

Percentage increase or decrease (property appreciation or depreciation) can be expressed in terms of a common ratio. The total increase can be found by multiplying by a common ratio.

E.g.#1 The population of Fredericton is 40,000. If the city expects a 10% increase by next year what will the new population be?

Method #1: Old way10% of 40,000 = 4000 Take 10% of 40,00040,000 + 4000 = 44,000 Add this answer to 40,000

The new population will be 44 000.

Method #2: New Way (Growth factor)This may be done another way. If we add 10% to 100% we have:

100%+10%110% Change this to decimal format by dividing by 100 = (1.10)

(1.10) is a growth factor. We use this to multiply by 40,000 so using the general formula we can substitute for a and b as follows:

y = a*bx where a = 40,000 initial populationb= (1.10) growth factor which represents 110%x = 1 year’s growth

y = 40 000*(1.10)1

y = 44 000

The new population will be 44 000

Exponential Growth

E.g.#2 A $20,000 car depreciates by 18% per year after you buy it. What will the car be worth in 5 years?

Since this is a depreciation question instead of adding the percent to 100% be instead subtract as follows:

100% -18% 82% Change this to decimal format by dividing by 100 = (.82)

(.82) is the decay factor. We use this to multiply by $20 000 so using the general formula we can substitute for a and b as follows:

y = a*bx where a = 20,000 initial car valueb= (.82) decay factor which represents 82%x = 5 year’s growth

y = 20 000*(.82)5

y = 20 000*(0.370739843)y = $7414.80

Asn(E3) Focus B Q 22 to 28 p.116 Q 30,31, 33 p. 118 Q. 10, 11 p. 113

Exponential GrowthAsn(E4) Investigation #3 Q 34 to 37 p. 119

Note(57) For exponential curves the growth factor or decay factor is affected by the sign of the variable. Ie. when x > 0 vs when x < 0(For each case below “a” = 1)

a) if x > 0 ie. x = [1,2,3,4, …] y >1

y = 2x let x = 1y = 21 +1y = 2

y = 2x let x = 2 x > 0 y = 22

y = 4

Observation: All y values will be greater that 1. (Never negative)I.e. the range is R = [y y >1, yR]

b) if x< 0 ie. x […,-2,-1]

y = 2x let x = -1y = 2-1

y = 1 2 0 < y < +1

+1y = 2x let x = -2y = 2-2

y = 1 x < 0 22

y = 1 4Observation: All y values will be less than 1 but greater than 0. (Never negative).Ie. the range is R= [y 0 < y < 1, y R]

Exponential Growth

Asn(E5) Investigation #4 Q 41 to 49 p. 122

Note(58) The value of “b” for the general formula y = a*bx determines the rate of growth (or decay).

Compare: y1 = 5x and y1 = 6x when x < 0

1) The common ratio for y = 5x is 5 so that b = 5. This means the y values or the graph will increase at a rate of 5 times for each increment of 1 on the x variable.

2) The common ratio for y = 6x is 6 so that b = 6. This means the y values or the graph will increase at a rate of 6 times for each increment of 1 on the x variable.

x -3 -2 -1 0y1 .008 .04 .20 Growing by a factor of 5y2 .0046 .028 .17 Growing by a factor of 6 (faster rate)

Conclusion: Even though the values for y2 are smaller than y1 the growth rate is still faster. When x = 0 the values of y2 began to surpass those of y1

Compare: y1 = 5x and y1 = 6x when x ≥ 0

x 0 1 2 3y1 1 5 25 Growing by a factor of 5y2 1 6 36 Growing by a factor of 6 (faster rate)

Conclusion: The rate of growth is still faster for the graph y = 6x compared to the graph y = 5x . (y2= y1 when x = 0) . When x >0 the y2 values are greater than the y1 values.

Exponential Growth

Note(59) The rules for exponent operations:

The product rule bm bn = bm+n

The quotient rule b m = bm-n

bn

The zero exponent b0 = 1

The negative exponent b-m = 1 and 1 = bn

bm b-n

The power of a product (ab)m = ambm

The power of a quotient a n = a n b bn

The power of a power (am)n = amn

Note(60) Success with evaluating powers is related to the understanding of what the base is in each power.

E.G. Determine the base of each power then evaluate.

E.g.#1 Meaning The base:

(2)4 = 2 2 2 2 = +16 The base is 2(-2)4 = (-2)(-2)(-2)(-2) = +16 The base is –2-24 = - 2 2 2 2 = -16 The base is 2. The subtract sign is not

affected by the exponent 4.E.g.#2 Answer The base:20 = 1 The base is 2(-2)0 = 1 The base is –2. The minus is affected by the 0 exp.-20 = -1 The base is 2. The minus is independent of 0 exp.

Asn(E6) Handout. (see next page of this printout)

Exponential Growth Asn(E6)

1.Simplify to a single base:a) 32 35

b) 36 33

c) (24)3

2. Simplify to a single base:a) (-3)12

(-3)4 (-3)6

b) 25 2-4

2-2 23

3. Simplify and write as a single power:a) (75 73 )2

b) (9-3 98)-2

4. Simplify these expressions:a) w -4 w6

b) n-11 n -4

c) x -4 y -6 x -5 y4

d) 24 x 3 y -2 12 x2 y3

e) (9a-1b0)(5a4b-3)

5. Evaluate: (No decimal answers)a) -30

b) (-4)0

c) -2-2

d) -(-3)-2

e) 1 5-2

6. Simplify and substitute for a = -2 and b = 3a) 18a -3 b 12

a-5b14

Math 112 Exponential GrowthNote(61) Algebraic exponents are handled the same way as regular exponents except you are required to group like terms (tiles) then simplify.

E.g. Simplify to a single base m:

(m 2a +b)(m a+3b) = m 2a+b+a+3b Add the exponents = m 2a+a +b+3b Group like terms

= m 3a +4b Add like terms

Asn(E7) Q22 a) to r) p. 169

Note(62) Fractional exponents may be written in root form as they are equivalent.Eg. 41/2 = 2 Evaluate on calculator 4^(1/2) = 2

4 = 2 Evaluate on calculator

The two powers above are equivalent so that we can say:41/2 = 4

Eg. Write each in root form and check with calculator91/2 = 9 = 3161/2 = 16 = 4251/2 = 25 = 5

In general: b1/ n = n b

Note(63) Cube roots appear in the form b 1/3 and the interpretation of this is that we are looking for the 3rd root of the base.E.g. Find the cube roots of each of the following:

Meaning:81/3 = 38 = Find 3 identical factors that when multiplied, give 8.

(2)(2)(2) = 8. The answer is 2 so that 38 = 2

271/3 = 327 = Find 3 identical factors that when multiplied, give 27.(3)(3)(3) = 27. The answer is 3 so that 327 = 3

641/3 = 364 = Find 3 identical factors that when multiplied give 64(4)(4)(4) = 64. The answer is 4 so that 364 = 4

Math 112 Exponential GrowthNote(64) The fourth root is similar to the 3rd root and is exemplified below:

161/4 = 416 = Find 4 identical factors that when multiplied give 16(2)(2)(2)(2) = 16. The answer is 2 so that 416 = 2.

Note(65) Rational exponents that have numerators and denominators may be written in radical form as well.

E.g. Evaluate the following by calculator:

8 2/3 = 4 Enter as 8 ^ (2/3)(38)2 = 4 Evaluate 38 by hand first then square the answer.

From the above we discover that the two are equivalent so that:8 2/3 = (38)2

Write the following in radical form then evaluate:16 3/4 = (416)3 Evaluate inside bracket to get 2 so (2)3 = 832 4/5 = (532)4 Evaluate inside bracket to get 2 so (2)4 = 1664 2/3 = (364)2 Evaluate inside bracket to get 4 so (4)2 = 16

Note(66) Negative rational exponents require that the reciprocal be taken first, then the base rewritten in radical form.

E.g. Evaluate the following without technology

16 -3/4 = 1 Take the reciprocal of the base 16 16 3/4

= 1 Write the denominator in radical form (416)3

= 1 Take the fourth root of 16 inside the brackets. (2)3

= 1 Cube the base 2. 8In general: b -m/n = 1

(n b )m

Asn(E8) Q 1,3,5 p. 164 Asn(E9) Q 7 to 10 p.164,165 Asn(E10) Q 54 p.123Test # 5


Note(67) The vertical stretch of an exponential function is determined by the value of “a” in the general exponential function y = a(b)x + c

e.g. y = 3(2)x a = 3 The vertical stretch is 3

This may be written in transformation form as 1/3y = (2)x

Note(68) The vertical translation of an exponential function is determined by the value of “c” in the general exponential function y = a(b)x + c

e.g. y = 3(2)x + 1 c = 1 The vertical translation is 1 up

This may be written in transformation form as 1/3( y -1) = (2)x

Special note: The imaginary line from which the function starts is y = 1. this line is called an asymptote.

Further note: The y- intercept is found 3 units up from this asymptote so that the y-intercept is given by adding a to c. So 3+1= 4 is the y-intercept.

y-intercept = “a + c” the asymptote has an equation defined by y = “c”


Note(69) The horizontal stretch of an exponential function is determined by the incremental change (factor) acting upon the x variable.

For the exponential function y =a(b)x/x + c x is the incremental change with the variable x.

e.g. y = (2)x/3 x = 3 The horizontal stretch upon x is 3 units.

Comparing y = (2)x to y = (2)x/3 we can see the graph is stretched horizontally by a factor of “3” units.

X 0 1 2 3 4 5 6y1 1 2 4 8 16 32 64y2 1 1.26 1.58 2 2.52 3.17 4

(x,y1) (x,y2)(1,2) (3,2) x is stretched by a factor of 3(2,4) (6,4) x is stretched by a factor of 3

3


Note(70) The horizontal translation of an exponential function is determined by the constant associated with the variable x.

For the exponential function given by y = a(b)x + d + c the horizontal translation is given by the variable “d” .

e.g. y = 1(2)x -3 d= -3 The horizontal translation is 3 units to the right. Comparing the tables of values for y = 1(2)x and y = 1(2)x-3 we can see that each point of the graph is translated horizontally 3 units right.

x 0 1 2 3 4 5 6y1 1 2 4 8 16 32 64

y2 .1251 .25 .5 1 2 4 8

(x,y1) (x,y2)(0,1) (3,1) x is translated 3 units right(1,2) (4,2) x is translated 3 units right(2,4) (5,4) x is translated 3 units right +3


Note(71) The reflection about the x-axis is given by the sign in front of the variable “a”.

e.g. y = -2(3)x + 4 The graph is a reflection on the x-axis.

y = -2(3)x + 4 is the mirror image of the graph defined by…..y = 2(3)x + 4 and the image line is the asymptote defined by y = 4

Note(72) The reflection about the y-axis is given by the sign in front of the variable x.

e.g. y = 2(3)-x + 4 The graph is a reflection on the y-axis

The function may be rewritten as y = 2(1/3)x + 4 and from this we can see that it is a decay curve. We conclude that the mirror image of a growth curve around the y-axis is a decay curve and vice versa.


Note(73) Vertical Stretch Questions. When c= 0 in the expression y = a(b)x + c both the vertical stretch and the y-intercept are given by the value of “a”

Asn(F1) Investigation #5 A , Q 1 to 5 p. 128 Q 7 p. 129 Q 9 to 12 p. 129,130.

An exponential function can be generated from a table of values or from the information given in a question.

Table of values: Look for the y intercept in the table or the “initial” or starting value in the question to find “a”. Divide successive terms to find the value of “b”.

Asn(F2) Q 19 to 23 p. 132,133 Advanced Q 24,25 p.133,134


Note(74) Horizontal Stretch Questions. For exponential functions, it is often common to ask “When will the value of a treasured object double or triple in price?”E.g. The exponential function y = 1.50(1.15)x generates the following table:

x1 0 1 2 3 4 5 6 7 8 9 10y 1.5 1.73 1.98 2.28 2.62 3.00 3.47 3.99 4.59 5.27 6.00

From this table we observe that the initial cost of $1.50 for a treasured toy (e.g. model car) doubles every 5 years. We may redo the chart in increments of 5 years as follows: (We can obtain this same result on the graphing calculator (set Δtbl = 5)

x1 0 5 10y 1.5 3.00 6.00

From this new table we can obtain values for “a” and “b” quite readily. “a” = 1.50 the starting value or y-intercept. “b” = 2 the common ratio obtained by dividing successive terms. Putting this all together we come up with an equivalent exponential function given by:y = 1.50(2)x/5 where: a = 1.50 b = 2 and x = 5Comparing we see that the two functions are equivalent so that:1.50(2)x/5 = 1.50(1.15)x

Asn(F3) Q 26 to 30 p. 135 to 136

Asn(F4) Q 31 to 37 p. 136 to 138 Advanced Q 38 p. 138


Note(75) Vertical Translation Questions. When c 0 in the expression y=a(b)x/x + c the graph is vertically translated up or down by the value of “c”. The asymptote of an exponential function is given by the letter c in the general expression:

y = a*bx +c c = location of the horizontal asymptote

E.g. For the function y = 3(2)x + 1 a = 3b = 2c = 1

From the graphing calculator the graph appears as follows:

+4(0,4) the y intercept is 4 not 3

+1 y = +1 equation of horizontal asymptote x-axis

We observe that the entire graph is vertically translated upward by 1 unit. The y-intercept is “4” not “3” so that in general the y-intercept is given by: a + cexample: a = 3 and c = 1 for the function y = 3(2)x + 1 so that:a + c =3 + 1 = 4

Asn(F5) Q 43 to 46 p. 140,141 Advanced Q 1,2,3,4 p. 144; Q 6 to 16 p. 149,150

Transformation form of an exponential function is given by:1/a(y – k) = b^ x/x -h Where:VS = a HS = xVT = +k HT = +h


Note(76) Exponential equations whose bases are equivalent may be solved by equating the exponents.

Eg.#1 Solve for x23 = 2x Bases are equivalent so exponents must be equivalent3 = x

E.g.#2 Solve for x34 = 32x Bases are equivalent so exponents must be equivalent 4 = 2x

4 = 2x Divide by 2 then cancel or reduce right side.2 22 = x

E.g. #3 Solve for x52 = 5 x/5 Bases are equivalent so exponents are equivalent2 = x

552 = 5 x Multiplication by 5 then cancel or reduce right side

5 10 = x

Note(77) Exponential equations whose bases are not equivalent need to have their bases reduced or rewritten so that they are equal.

E.g. #4 Solve for x8 = 2x Rewrite 8 as base 223 = 2x Equate exponents3 = x

E.g. #5 Solve for x16 = 1 x Rewrite 16 as base 2

224 = 1 x Rewrite ½ as base 2 with a negative exponent 224 = (2-1)x Carry –1 through brackets to outside by multiplication24 = 2–x Equate the exponents4 = -x Write the opposite (Flip tiles both sides)

-4 = x

Math 112 Exponential GrowthE.g. #6 Solve for x

1 = (9)x Write 27 as base 3271 = (9)x Write fraction as base with negative exponent33

3-3 = (9)x Write 9 as base 33-3 = (32)x Multiply exponents 2 and x3-3 = 32x Equate exponents-3 = 2x Divide both sides by 2-3 = 2x Cancel or reduce right side 2 2-3 = x 2

Note(78) Exponential equations may involve the variable with more than one base. We follow the normal steps of equation solving.E.g. Solve for x

22x +1= 2x+3 Equate exponents2x+1 = x +3 Move variable to left. Subtract x2x-x +1 = x –x +3 Simplifyx + 1 = 3 Move constant to right. Subtract 1x +1 –1 = 3 –1 Simplifyx = 2

Note(79) Exponential equations in the form y = a*bx + c require several steps prior to equating the exponents.E.g. Solve for x 70 = 12(6)-x+2 – 2 Isolate base 6 by adding 2 to both sides70 +2 = 12(6)-x+2 – 2 +2 Simplify 72 = 12(6)-x+2 Divide by 12 72 = 12(6)-x+2 Cancel or reduce 12 12 61 = (6)-x+2 Equate exponents 1 = -x +2 Subtract 2 from both sides -1 = -x+2 Write the opposite or flip both sides 1 = xAsn(F6) Q 12,13,14,16 p. 160Asn(F7) Q 8,9,15,17,18,10,11 p. 158 to 161 Q. 19 p. 161


Note(80) Unknown bases for exponential equations can be solved by multiplying by the reciprocal of the exponent.

E.g. Solve for a

a3/2 = 64 Multiply both bases by the reciprocal 2/3a 3/22/3 = 64 2/3 Cancel or reduce exponents with base aa = (362)2 Take cube root of 64a = (4)2 Square base 4a = 16

E.g. Solve for bb –2/3 = 16 Multiply both bases by the reciprocal –3/2

b –2/3-3/2 = 16 –3/2 Cancel or reduce exponents with base bb = 16 –3/2 Take reciprocal of base 16b = 1 Take square root of base 16 (216) 3

b = 1 Cube base 4 (4)3

b = 1 64


Note(81) Radioactive decay is a special case involving the exponential function. The language associated with radioactive decay requires interpretation.

E.g. The half-life of radium is 1620 years. The initial mass of the radium is 10 g. (The half-life of radium is the time it takes to lose ½ of its mass). The common ratio is therefore ½ or the factor, which is used for the decay curve.

a) How much radium will be present in 2500 years?

a = 10 the initial mass b = ½Δx = 1620 the change in years for each half-lifex = 2500 (time)

Substitution:y = a*b x/Δx

y = 10*(1/2)x/1620 Substitute x = 2500y = 10*(1/2)2500/1620 Evaluate on calculatory = 3.4 g 3.4 grams of the original 10 grams

will be present in 2500 years.

b) When will the radium be 1/5 of its original mass?1/5 of the original mass is 1/5 10 g = 2 grams. We substitute into the equation in place of y and solve for x.

y = a*b xΔx

2 = 10*(1/2)x/1620

Step 1: Enter right side into the graphing calculator after \y1 = Step 2: 2nd tableset. Set: tblstart to 0 and Δtbl to 1000 since Δtbl is 1620Step 3: Scroll down until the y value is approximately 2Step 4: Reset Δtbl to 10 and scroll to find xx = 3760


Note(82) The range of exponential functions are limited by the asymptote of the function so that:

R = {y y > c, y R} where c comes from y = a*bx +c

E.g. Find the range and domain of these functions:

y = 2x +1 y = 3x –2

+1

-2

D = {x x R} D = {x x R}R = {y y >1, y R} R = {y y > -2, y R}

+3

D = {x x ≥ 0,x R}R = {y y ≥+3, y R}Asn(F8) Advanced Q 22,23,24,26,27 p. 153 to 155Remember to write in full transformation form before identifying transformations. ½(y-3) = (2) –x/4 -2 = ½(y-3) = (2) -1/4(x +8) Test #6


Note(83) An inverse function is a function that when applied to the Range values produced by the original function returns the original Domain values from which they came.. The inverse function returns the original values of the initial function.

e.g.original function:y = x2 let x = 5y = (5)2

y = 25so y = 25

inverse function:x = y let y = 25x = 25x = 5 so x = 5 We have the original value back of x = 5

Because the final value 5 is the same as the initial value 5 we say that the function x = y is the inverse function of the original function y = x2

Note(84) The inverse function of an exponential function is the logarithmic function.e.g.original function:y = 10x let x = 3y = 103

y = 1000 so y = 1000

inverse function:x = log10(y) let y = 1000x = log10(1000)x = 3 so x = 3

Because the final value 3 is the same as the initial value 3 we say that the function x = log10(y) is the inverse of the original function y = 10x.

Asn(G1) Q 1 to 10 p. 172 to 174

Math 112 Exponential GrowthNote(85) Inverse functions are symmetrical about the line y = x. A table of values for the function y = 10x and its inverse y = Log2(x) are given below:

Inverse (switch variables)y = 10x x = 10 y or y = log 10 (x)

x y x y1 10 10 1 2 100 100 23 1000 1000 3

(Show that the y values in second table can also be found by applying the function y = log 10 (x) to x values in second table. This will prove that x = 10 y can be rewritten as y = log 10 (x ) )

+8 y = 10x (original function)+7 y = x (axis of symmetry)+6+5+4+3 x = 10 y can be written as…+2 y = log10 (x) (inverse function)+1

1 2 3 4 5 6 7 8

Note: The above graph can be illustrated on the graphing calculator.

Note(86) From the above diagram and tables we conclude that functions in the form 10y = x can be written in the form log10 (x) = y. A “proof” is given below by evaluation on the calculator.

E.g. if 103 = 1000 then log10(1000) = 3 (evaluate on calculator) if 104 = 10000 then log10(10000) = 4 if 10y = x then log10(x) = yIn general: if be = a then logb(a) = e Example: if 23 = 8 then log28 = 3

Asn(G2) Q 11 to 15 p. 174,175 Advanced Q 16,17 p. 175

Math 112 Exponential GrowthNote(87) Logarithm of a product

If we enter the following on your calculator we obtain the result to the right.log(23) = .778log(2) + log(3) = .788

We generalize as follows: (Other examples may be done)log(23) = log(2) + log(3)

In general:log(mn) = log(m) + log(n)

Note(88) Logarithm of quotient

If we enter the following on your calculator we obtain the result to the right.log(42) = .301log(4) - log(2) = .301

We generalize as follows: (Other examples may be done)log(42) = log(4) - log(2)

In general:log(mn) = log(m) - log(n)

Note(89) Logarithm of a power

If we enter the following on your calculator we obtain the result to the right.log(42) = 1.2042log(4) = 1.204

We generalize as follows: (Other examples may be done)log(42) = 2log(4)

In general:log(mn) = n*log(m)

Asn(G3) Q 1 to 3 p. 177,178 Advanced Q 7,8 p. 178Math 112 Exponential Growth

Note(90) Solving exponential equations using logarithms.

E.g. Solve for x

9x = 27 Apply log10 to both sideslog10(9x) = log10(27) Apply power log rule to left side onlyx log10(9) = log10(27)x log10(9) = log10(27) Divide both sides by log10(9) log10(9) log10(9) x = (1.4314) Evaluate (.9542)

x = 1.5

Therefore we can shorten the original equation in a few steps as follows:

9x = 27x = log10(27) log10(9)x = 1.5

In general:If be = a then e = log10(a)

log10(b)

Asn(G3) cont’d Q 4,5,6 p. 177,178


Note(91) Log identities can be derived by equating two equal expressions for an exponential equation.

E.g. 9x = 27 Solving for x we have: x = log10(27) log10 (9)

9x = 27 In logarithmic form we have: x = log9(27)

Since we have two expressions for x, we may equate the two as:

log9(27) = log10(27) = 1.5 log10(9)In general:Logb(a) = log10(a)

log10(b)

Special Note: The significance of the above rule cannot be overemphasized as it indicates that if a log is not written in base 10 it can be rewritten in base 10 and evaluated on any calculator.

SUMMARY OF LOG RULES AND IDENTITIES

1. If be = a then logb(a) = e Exponent form to Log form

2. logb(a) = log10(a) Calculator log form log10(b)

3. If be = a then e = log10 (a) Exponent form to calculator form log10(b)4. log(mn) = log(m) + log(n) Log of a product

5. log(mn) = log(m) - log(n) Log of a quotient

6. log(mn) = n*log(m) Log of a power

Asn(G4) Q 9 to 11 p. 180,181 Q 12 in reverse order.


Note(92) Problem solving requires interpretation and proper application of logarithm identities.E.g#1. A bacteria colony’s original population is 1000. It doubles every hour. When will it reach a population of 5000?y = a*bx/Δx a = 1000 original population

b = 2 doublesΔx = 1 every houry = 5000 future population

Substituting we have:5000 = 1000(2)x Solve for x5000 = 1000(2)x Isolate base 2 by dividing both sides by 10001000 1000

5 = (2)x Rewrite as log base 10 x = log10(5) Evaluate right side log10(2) x = 2.3 hours

E.g.#2 A rabbit population triples every 4 years. If there were 435 rabbits to start with, when will it reach a population of 6781 rabbits?y = a*bx/Δx a = 435 to start with

b = 3 triplesΔx = 4 every 4 yearsy = 6781 future population

Substituting we have:6781 = 435(3)x/4 Divide by 435 both sides6781 = 435(3)x/4 Reduce both sides 435 43515.59 = (3)x/4 Rewrite as log base 10 x/4 = log10(15.59) Evaluate right side log10(3) x/4 = 2.5 Multiply both sides by 4 x = 10 years

Asn(G5) Handout Q 1,3,4,5 Advanced 15,16, 18 to 26 p. 182,183Test #7

Math 112 Trigonometry Book 2 Mathematical Modelling (old version-blue text) required

Note (90) Trigonometry involves the study of angles and sides in relation to triangles in particular. A review of terms is essential

1. complementary angles- angles that add up to 900.2. supplementary angles- angles that add up to 1800.3. acute angle- an angle that is less than 900.4. obtuse angle- an angle that is between 900 and 1800.5. right angle- a angle with measure exactly 900.6. straight angle- an angle with measure exactly 1800.7. isosceles triangle- a triangle with two sides equal.8. equilateral triangle- a triangle with all sides equal.9. right angle triangle- a triangle with one angle of measure 900.10. acute triangle- a triangle with all angles less than 900

11. obtuse triangle- a triangle with one angle greater than 900

12. oblique triangle- any type of triangle that does not have a 900

(both the acute and obtuse triangles are oblique).

Note(91) Right angle triangles are special triangles in that the sides are related in a special manner.

Pythagorean Property:

A c2 = a2 + b2

Or H2 = S12 + S2

2

b c

C a B

Math 112 Trigonometry

Note(92) Right angle triangles are special triangles in that the angles are related in a special manner.

sum = 1800

A A +B +C= 1800.

Because C is 900 we have a special b c relationship between A and B as follows:

A +B = 900

C a B

Note(93) Some triangles cannot be solved using the Pythagorean Theorem or the angle sum of a triangle. Primary trigonometric ratios can be used instead.

E.G. solve for x in the figure below:

A

12

350

B x C

In such a case, trig ratios can be used for solving for the unknown. The primary trig ratios are:

Sin() = Opp We use an acronym for assisting in the recall Hyp of these ratios:

SOH CAH TOACos() = Adj Hyp

Tan() = Opp Adj


Note(94) We may use the primary trig ratios in solving for the sides of a right triangle as illustrated below:

E.g. Solve for x in the figure below:

1) Shade the angle given12 2) Using the angle as a point of reference

x label the side x as opposite and 12 as the hypotenuse.

300 3) Choose the trig ratio that fits the situation.

Sin() = Opp Write the primary trig ratio that matches HypSin(300) = x Substitute for and Hyp 12 12Sin(300) = x Multiply both sides by 12. (cross multiply)

x = 6

E.g. Solve for x in the figure below:

x 10

450

Sin() = Opp Write the primary trig ratio that matches HypSin(450) = 10 Substitute for and Opp. xx * sin(450) = 10 Cross multiply to move x to left side.x * (.707106..) = 10 Divide both sides by (.707106…)x = 14.1 cm

Asn(H1) Question involving primary trig ratio and sides. (In class)


Note(95) We may use the primary trig ratios in solving for the angles of a right triangle as illustrated below:

E.g. Solve for 1) Shade the indicated angle2) Label the sides relative to the shaded angle3) Choose the trig ratio that matches the situation

4

5

Tan() = Opp Write the primary trig ratio that matches AdjTan() = 4 Evaluate the right side to obtain a decimal. 5Tan() = .8000 Use 2nd function tan-1 to find angle

= tan-1 (.8000)

= 38.70

Asn(H2) Questions involving primary trig ratios and angles. (In class)


Note(96) There are 6 parts that make up a right triangle. (3 sides and 3 angles). We may solve for the unknown parts in a variety of ways:

Case #1 the triangle has 2 angles given and one side.Suggestion: Start by solving for the 3rd angle

Tools to use for solving triangles: x Tool #1 Angle sum of triangle2 Tool #2 Trig ratios

Tool #3 Pythagorean Theorem 350

y

Tool#1 sum of triangle: + 350 + 900 = 1800. = 550

Tool#2 Trig ratios:Tan(350) = Opp Write the trig ratio that matches the question AdjTan(350) = 2 Substitute y

y*tan(350) = 2 Multiply both sides by y. (Cross multiply)y = 2/tan(350) Divide both sides by tan(350)

y = 2.85 Evaluate

Tool#2 Trig ratios:Sin(350) = Opp Write the trig ratio that matches the question Hyp

Sin(350) = 2 Substitute xx*sin(350) = 2 Multiply both sides by x. (Cross multiply)

x = 2/sin(350) Divide both sides by sin(350)x = 3.48 Evaluate


Case #2 the triangle has 2 sides given and one angle.Suggestion: Start by solving for the 3rd side

Tools to use for solving triangles: 14 Tool #1 Angle sum of trianglex Tool #2 Trig ratios

Tool #3 Pythagorean Theorem

5Tool#3 Pythagorean Theorem:c2 = a2 + b2

(14)2 = (x)2 + (5)2 Substitute for c and b196 = x2 + 25 Square termsx2 = 196-25 Isolate xx2 = 171 Simplifyx = 171 Take sq rt of 171

Tool#2 Trig ratios:Cos() = Adj Write the trig ratio that matches the question HypCos() = 5 Substitute for adj and hyp 14Cos() = .35714… Evaluate right hand side

= cos-1(.35714..) Use 2nd function cos-1

= 690

Tool#2 Trig ratios:Sin() = Opp Write the trig ratio that matches the question HypSin() = 5 Substitute for opp and hyp 14Sin() = .35714… Evaluate right hand side

= sin-1(.35714..) Use 2nd function sin-1

= 210

Asn(H3) Handout to do at home.


Note(97) Application of trigonometry to real life situations requires an understanding of common terms.

Angle of elevation- the angle between the ground and the line of sight looking upwards.

Angle of depression- the angle between a line horizontal to the ground and the line of sight looking downwards

Special note: the angle of depression equals the angle of elevation.

E.g. The angle of depression from the top of a cliff to a sailboat in the bay below is 250. If the sailboat is 100 m from the shore how high is the cliff.

250

angle of depression 250

x angle of elevation is also 250

angle of elevation sailboat 100 m

The angle of elevation is equal to the angle of depression since they are alternate angles between parallel lines.

Tan(250) = x 100

x = 46.6 m

Math 112 TrigonometryNote(98) The area of a triangle is given by the formula: A = ½ b*h.The area may also be found by using the formula A = ½ bcSin(A) where A is the included angle.

E.g. Find the area of the Δ ABC given below:

A

c = 5.4 h

B 18.40

D C a = 9.3

The area of Δ ABC can be found using the formula:A = ½ b*hA = ½ 9.3*h

Using the triangle Δ ABD and sine trig ratio we find h as follows:Sin(18.4) = Opp = h

Hyp 5.4 h = 5.4 Sin(18.4) h = 1.7

Substituting h = 1.7 into the area formula above we have the area as:A= ½ (9.3)(1.7)A = 7.93

The area of Δ ABC can also be found using the formula:A = ½ ac*Sin(B) = ½ (9.3 )(5.4 ) sin(18.4)A = 7.93

Conclusion: The area formula using the Sine trig ratio is given by:A = ½ acSin(B) where B is the included angle between sides a and c.

Asn(H4) Handout on area problems.


Note(99) The sine law can be used for solving oblique triangles. We solve for either an unknown side or an unknown angle.

E.g. #1 Solve for side “a” in the figure given below.

A 400

b = 12

350

B a = ? C

The sine law for sides is given by: a = b = c Sin(A) Sin(B) Sin(C)

Substituting we have:

a = (12 ) = c Since c and Sin(C ) are not known Sin(40) Sin(35) Sin(C) we do not use the last ratio for

evaluation.Evaluate: a = (12 ) Cross multiply to isolate side “a”Sin(40) Sin(35)

a = 12 Sin(40) Evaluate Sin(35)

a = 13.4


E.g. #2 Solve for angle “B” in the figure given below.

A 950

c = 16 b = 13

B a = 14 C

The sine law for angles is given by:

Sin(A) = Sin(B) = Sin(C) The ratios are reciprocated a b c for easier evaluation.

Substituting we have:

Sin(95) = Sin(B) = Sin(C) Since we are not solving for C we (14) (13) (16) do not use the last ratio to evaluate.

Evaluate:

Sin(95) = Sin(B) Cross multiply to isolate Sin(B) (14) (13)

13Sin(95) = Sin(B) Evaluate left side 14

0.9250 = Sin(B) Apply 2nd function Sin-1

Sin-1(0.9250) = B Evaluate

B = 67.70

Asn(H5) Q 1 to 3 p. 259


Special Note: We use the Sine Law if the angles and sides are opposites. We look for the “butterfly- pattern” as illustrated below;

A 950

c = 16 b = 13

B a = 14 C

Special Note: Occasionally a required angle is missing in which case we can utilize the angle sum of a triangle to solve for the missing angle.

Solve for “b”A 300

c = 12 b = ?

500 ? Missing angle ( sum gives 1000) B C

We cannot use the Sine Law until we find the size of the unknown (?). To do this we use the angle sum of a triangle:

sum = 1800 so:A +B +C = 1800

30 + 50 + C = 180 C = 100

We can now use the Sine Law to solve for side b.Asn(H6) Q 4a) b) 5 p. 259Asn(H7) Handout Q 5,6,7 Sine Law Problems

Math 112 TrigonometryNote(100) The cosine law can also be used for solving oblique triangles where the sine law is not helpful.

E.g.#1 Solve for side “a”

C

a = ? b =12

B 620 A c = 10

In the case above , we are lacking information that is required for the sine law, that is a second angle.

We use another law designated as the cosine law given by:

a2 = b2 + c2 –2bcCos(A) where A is included between two sides a,bb2 = a2 + c2 –2acCos(B) where B is included between two sides a,cc2 = a2 + b2 –2abCos(C)where C is included between two sides b,c

Using the first of these from above we have:

a2 = b2 + c2 –2bcCos(A) Substitutea2 = (12)2 + (10)2 –2(12)(10)Cos(62) Enter into graphing calculatora2 = 131.33a = 131.33 Apply square root to both sidesa = 11.5

Math 112 TrigonometryNote(100) The cosine law cont’d

E.g.#2 Solve for angle “A”

The following rearranged cosine laws are useful in solving for angles:

1) Cos(A) = b 2 + c 2 - a 2 2bc2) Cos(B) = a 2 + c 2 - b 2 2ac3) Cos(C) = a 2 + b 2 - c 2 2ab

C

b = 15 a = 12

A B c = 16

Cos(A) = b 2 + c 2 - a 2 2bcCos(A) = (15) 2 + (16) 2 - (12) 2 Substitute 2(15)(16) Cos(A) = .70208…. Evaluate A = cos-1(.70208….) Apply second function cosine A = 45.40

Asn(H8) Q 8,9,10,12,13,15 p. 261,262 Cosine Law problemsAsn(H9) Handout cosine law practice sheet. (Optional)Asn(H10) Handout- problem solving. Sine, Cosine and Area questions.

Advanced Q 11, 14 p. 262 Q 21 p 264 q.24, 25. p. 265


Note(101) The ambiguous case for the sine law.

Some geometric constructions allow for 2 possible constructions from the given data. Two correct results are possible.

E.g. Construct 2 unique triangles using the following data:A = 270 b = 12.3 cm a = 9.6 cm

Construction #1 C

b = 12.3 a = 9.6

A 270 B In the above diagram B = 35.60 if measured with a protractor. B is acute.

Construciton #2 C

b = 12.3 a = 9.6

270

A B

In the above diagram B = 144.40 if measured with a protractor.B is obtuse.

Since both diagrams are correct then there are 2 possible answers for B as follows:

B = 35.60

B = 144.40

It is noteworthy that the measures 35.60 and 144.40 are supplementary.


Note(102) For the sine function we notice that the sine ratios for supplementary angles are identical.

E.g. Sine(35.6) = .5821229 = Sine(144.4)

This is true for all supplementary angles:Sine(10) = .1736.. = Sine(170)Sine(30) = .5000 = Sine(150)

In general:Sine() = Sine(180 - )

Note(103) If given an obtuse angled triangle, it is possible to obtain a result that does not match the diagram. In such a case we suggest that the supplementary angle is the correct result.

E.g. Solve for B in the figure below:

b = 14 a = 10

300

A BSolve for B:

Sin( B) = Sin( A) b aSin(B) = 14Sin(30) 10Sin(B) = .7000 B = sin-1(.7000) B = 44.40 Because this does not match the appearance of the diagram we suggest that the supplement to this angle is the solution so that:

B = (1800 - 44.40)= 135.60

Asn(H11) Q 19,17 p. 264 Test # 8

math 112 quadratics notes - tranquills2 -...

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