http://www.youtube.com/watch?v=cqj5Qvxd5MO

Linear and Quadratic Functions and Modeling

A polynomial function is a function of the form:

on

nn

n axaxaxaxf 1

11

All of these coefficients are real numbers

the exponent must be a nonnegative integer

Remember integers are … –2, -1, 0, 1, 2 … (no decimals or fractions) so nonnegative integers would be 0, 1, 2 …

The degree of the polynomial is the largest power on any x term in the polynomial.

2

1

xx

Not a polynomial because of the square root since the power is NOT an integer

xxxf 42

Determine which of the following are polynomial functions. If the function is a polynomial, state its degree.

A polynomial of degree 4.

2xg

12 xxh

23x

xxf

A polynomial of degree 0.

We can write in an x0 since this = 1.

Not a polynomial because of the x in the denominator since the power is NOT nonnegative 11 x

x

x 0

Graphs of polynomials are smooth and continuous.

No sharp corners or cusps No gaps or holes---it can be drawn without lifting pencil from paper

This IS the graph of a polynomial

This IS NOT the graph of a polynomial

POLYNOMIAL FUNCTIONS OF NO AND LOW DEGREE

• NAME FORM DEGREE

• Zero function f(x) = 0 Undefined

• Constant function f(x) =a 0• Linear function f(x) ax + b 1• Quadratic function f(x) = ax² +bx + c 2• a ≠ 0

LINEAR EQUATIONS132 x

xx 2513

x421 43 x

In general, the equation of a horizontal line is y = b, where b is the y coordinate of any point on the line.In general, the equation of a horizontal line is y = b, where b is the y coordinate of any point on the line.

x

y

If we look at any points on this line we see that they all have a y coordinate of 3 and the x coordinate varies.

(-4, 3)

(-1, 3)

(2, 3)

Let's choose the points (-4, 3) and (2, 3) and compute the slope.

06

0

42

33

m

This makes sense because as you go from left to right on the line, you are not rising or falling (so zero slope).

The equation of this line is y = 3 since y is 3 everywhere along the line.

To find the average rate of change of a function between any two points on its graph, we calculate the slope of the line containing the two points.To find the average rate of change of a function between any two points on its graph, we calculate the slope of the line containing the two points.

(7, 8)

(0, 0)7

8

07

08

12

12

xx

yym

So the average rate of change of this function between the points (0, 0) and (7, 8) is 8/7

Slope of this line

S CAT T

ERD

I AGR

AMS

LINEAR CURVE FITTING

x

y

A scatter diagram is a plot of ordered pairs generally obtained by observation of some relation.

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

6 .0

7 .0

-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2 2 .5

0 .0

1 .0

2 .0

3 .0

4 .0

5 .0

6 .0

7 .0

8 .0

9 .0

1 0 .0

-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2 2 .5

-1 0 .0

-8 .0

-6 .0

-4 .0

-2 .0

0 .0

2 .0

4 .0

6 .0

8 .0

1 0 .0

-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2 2 .5

-1 2 .0

-1 0 .0

-8 .0

-6 .0

-4 .0

-2 .0

0 .0

2 .0

4 .0

6 .0

8 .0

-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2 2 .5

-1 5 .0

-1 0 .0

-5 .0

0 .0

5 .0

1 0 .0

1 5 .0

-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2 2 .5

-1 5 .0

-1 0 .0

-5 .0

0 .0

5 .0

1 0 .0

1 5 .0

-2 .5 -2 -1 .5 -1 -0 .5 0 0 .5 1 1 .5 2 2 .5

Look at these scatter diagrams to see if the relation looks linear or nonlinear.

LINEAR NONLINEAR LINEAR

NONLINEAR LINEAR NONLINEAR

15

17

19

21

23

25

27

29

50 52 54 56 58 60 62 64 66

speed of car

mile

s pe

r gal

lon

Let's look at some data gathered about the relationship between the speed of a certain car and the miles per gallon it getsspeed

xMPG

yordered pair

(x, y)50 28 (50, 28)52 28 (52, 28)54 27 (54,27)55 26 (55,26)55 25 (55,25)56 24 (56,24)58 24 (58,24)58 25 (58,25)60 22 (60,22)60 20 (60,20)62 20 (62,20)62 21 (62,21)63 20 (63,20)65 17 (65,17)65 15 (65,15)

We can plot these points and see if it looks like there is a relationship

It looks like a linear relationship because as you look it seems to have the pattern of a line with negative slope

We could come up with a function to estimate the miles per gallon given the speed. We'll pick two points on or near the line we made and find the slope and then use the point-slope formula.

15

17

19

21

23

25

27

29

50 52 54 56 58 60 62 64 66

speed of car

mile

s pe

r gal

lon

We'll choose (52, 28) and (65, 17)

(52, 28)

(65, 17)

13

11

5265

2817

12

12

xx

yym

11 xxmyy

5213

1128 xy

7213

11 xy

13727013

11y

Find the estimated MPG if the car speed is 70 mph

mpg

Click here for help on how to use your calculator to create a scatter diagram and find the line of best fit.

We could each come up with a slightly different line if we picked two different points to use. There is a process for finding the best line. This process is covered in a statistics class. We'll just use the calculator to find what is called the "line of best fit".

Properties of the Correlation Coefficient, r

• 1. -1 ≤ 1• 2. When r > 0, there is a positive linear

correlation• 3. When r < 0, there is a negative linear

correlation• 4. When | r | ≈ 0, there is weak or no

linear correlation.

2xxf

2

3

1xxf

23xxf

2

3

1xxf

2xxf 23xxf

If a > 0 the parabola opens up and the larger the a value the “narrower” the graph and the smaller the a value the “wider” the graph.

If a < 0 the parabola opens down and the larger the a the “narrower” the graph and the smaller the a the “wider” the graph.

2axxf

khxaxf 2

vertical shift, moves graph vertically by k

horizontal shift, moves graph horizontally by h

Determines whether the parabola opens up or down and how “wide” it is

162 xxxfWe need to algebraically manipulate this to look like the form above. We’ll do this by completing the square.

____1___62 xxxf

Add a number here to make a perfect square

Subtract it here to keep things equal (can’t add a number without compensating for it and we don’t want to add it to the other side because of function notation)

9 9

This will factor into (x-3)(x-3) so we can express it as something squared and combine the -1 and -9 on the end.

103 2 xxf

The graph of this function is a parabola

Let’s look at a quadratic function and see if we can graph it.

103 2 xxf

right 3

down 10

We started with and completed the square to get it in the format to be able to graph using transformations.

162 xxxf

We can take the general quadratic equation and do this to find a formula for the vertex. (This is done in your book). What we find from doing this is on the next slide.

cbxaxxf 2

The x value of the vertex of the parabola can be found by computing

a

b

2

The y value of the vertex of the parabola can be found by substituting the x value of the vertex in the function and finding the function value.

162 xxxfLet’s try this on the one we did before:

a

bx

2 vertex of value

1

(1)

(-6) 3

1013633 vertex of value 2 fy

The vertex is then at (3, -10)

(3, -10)

Let’s plot the vertex: Since the a value is positive, we know the parabola opens up.

The parabola will be symmetric about a vertical line through the vertex called the axis of symmetry.

162 xxxf

Let’s find the y intercept by plugging 0 in for x.

10602 xf

So y intercept is (0, -1)

The graph is symmetric with respect to the line x = 3 so we can find a reflective point on the other side of the axis of symmetry.

(0, -1) (6, -1)

We can now see enough to graph the parabola

(3, -10)

Let’s look at another way to graph the parabola starting with the vertex:

We could find the x intercepts of the graph by putting f(x) (which is the y value) = 0

162 xxxfThis won’t factor so we’ll have to use the quadratic formula.

160 2 xx

So x intercepts are (6.2, 0) and (- 0.2, 0)

a

acbbx

2

42

12

11466 2

2.0 and2.62

406

A mathematical model may lead to a quadratic function. Often, we are interested in where the function is at its minimum or its maximum. If the function is quadratic the graph will be a parabola so the minimum (if it opens up) will be at the vertex or the maximum (if it opens down) will be at the vertex.

We can find the x value of the vertex by computing a

b

2

We could then sub this value into the function to find its minimum or maximum value.

xpR

4000,2002

1 xxp

DEMAND EQUATION

The price p and the quantity x sold of a certain product obey the demand equation:

This is the real world domain. The equation doesn’t make sense if the quantity sold is

negative (x < 0) and it doesn't make sense if the price is

negative (if x > 400)

Express the revenue R as a function of x.

Revenue is the amount you bring in, so it would be how much you charge (the price p) times how many you sold (the quantity x)

xxxxR 2002

1200

2

1 2

xxR 2002

1 2 This is a quadratic equation and since a is negative, its graph is a parabola that opens down. It will have a maximum value then at the y value of the vertex.

What is the revenue if 100 units are sold?

1002001002

1 2 R 000,15$

What quantity x maximizes revenue?

Since the revenue function is maximum at the vertex, we'll want to find the x value of the vertex to answer this.

200

21

2

200

2

a

bx

What is the maximum revenue?

This would be the y value of the vertex

2002002002

1200 2 f

000,20$

4000,2002

1 xxp

DEMAND EQUATION

The price p and the quantity x sold of a certain product obey the demand equation:

What price should the company charge to receive maximum revenue?

xxR 2002

1 2

Since we just found that the quantity to achieve maximum revenue was 200, we can substitute this in the price equation to answer this question.

100$2002002

1p