fx x x 3 x b xba x a x b - university of north georgia · standard form): xh y k r 222 form reveals...

Hartfield – College Algebra (Version 2017a - © Thomas Hartfield) Unit THREE | Page - 1 - of 34

Topic 23: Average Rate of Change

Definition: The average rate of change of a function ( )y f x between x = a andx = b is

change in ( ) ( )avg rate of change = change in

y f b f ax b a

The average rate of change is equal to the slope of the secant line between x = a and x = b on the graph of f; that is, the line passes through

, ( ) and , ( ) .a f a b f b

Ex. 1 Find the average rate of change of fover the given interval.

37 1,2f x x x



37 2,3f x x x


3 1,4g x x x


Topic 24: The Difference Quotient

The concept of a derivative in calculus is based upon an algebraic concept called the difference

quotient: .f x h f x

h

In algebra, we can associate the difference quotient with the idea of average rate of change.

x represents the value of a while x + h represents b(and thus the the difference in b and a is h). The difference quotient allows you to quickly find an average rate of change for many different a and b.

The process of finding a difference quotient involves finding f(x + h), setting up the overall expression, and then simplifying the expression.

Ex. 1 Find the difference quotient for 3 4.f x x


Ex. 2 Find the difference quotient for 22 .f x x x

Ex. 3 Find the difference quotient for 2 4 .f x x x


Ex. 4 Find the difference quotient

for 1.1

xf xx


Topic 25: Review of Solving Quadratic Equations

Three methods to solve equations of the form 2 0.ax bx c

1. Factoring the expression and applying the Zero Product Property

2. Completing the square and applying the Property of Square Roots

3. Applying the Quadratic Formula

Ex. 1 Solve by factoring and applying the Zero Product Property.

2 2 48 0x x


Ex. 2 Solve by completing the square and applying the Property of Square Roots.

2 6 2 0x x

Ex. 3 Solve by applying the Quadratic Formula.

22 4 1 0x x2 4

2b b acx

a


Ex. 4 Solve by each of the previous two methods.

2 8 4 0x x


Topic 26: Solving Equations Based on Quadratics

The first set of equations are described as quadratic-type when an appropriate substitution is applied.

Ex. 1 Solve.

24 7 4 6 0x x

Ex. 2 Solve.

2 13 37 12 0x x


Ex. 3 Solve.

4 29 8 0x x

The second set of equations are actually radical equations, related to a previously covered type of equation, but the nature of the examples results in using quadratic equations. Caution: examples like these must be checked for false solutions!

Ex. 4 Solve.

3 5x x


Ex. 5 Solve.

2 5 3x x

Ex. 6 Solve.

4 2 5x x


Topic 27: Circles

Are both of these figures circles?

Definition: A circle is the set of all points equidistant to a fixed point.

The fixed point is called the center and the common distance from the center to each point of the circle is called the radius.


General Equation of a Circle:

2 2 0, where 0Ax By Cx Dy E A B

Ex. 1 2 2 6 4 12 0x y x y

No explicit information about circle conveyed.

Center-Radius Form of a Circle (also called Standard Form):

2 2 2x h y k r

Form reveals the center (h, k) and the radius r.

Steps to rewriting a circle equation into center-radius (standard) form:

1. Group like-lettered variables together and move the constant to the other side.

2a. Complete the square in each set of parentheses.

2b. Compensate for the values added into the parentheses.

3. Factor the groups and simplify both sides.


Ex. 1 Rewrite the equation into center-radius form (standard form), identify the coordinates of the center and the length of the radius, and sketch a graph on the coordinate plane.

2 2 6 4 12 0x y x y

Ex. 2 Rewrite the equation into center-radius form (standard form), identify the coordinates of the center and the length of the radius, and sketch a graph on the coordinate plane.

2 2 4 8 4 0x y x y


Ex. 3a Rewrite the equation into center-radius form (standard form), identify the coordinates of the center and the length of the radius, and sketch a graph on thecoordinate plane.

2 2 10 2 22 0x y x y

Ex. 3b Calculate the x-intercepts and y-intercepts of the circle.

2 2 10 2 22 0x y x y


Ex. 4 Find the standard equation of a circle from information.

passes through (4, 5)


A circle centered in the first quadrant with a radius of 4 that is tangential to both axes.



A circle with diameter endpoints of (9, 3) & ( 1, 3).


Topic 28: Quadratic Functions

Definition: A quadratic function is a function which can be written as

2 , where 0.f x ax bx c a

Properties of Quadratic Functions

1. The graph of a quadratic function is a parabola.If a is positive, the parabola opens up. If a is negative, the parabola opens down.

1. Parabolas have a turning point called a vertex.

2. Parabolas have an axis of symmetry which is a vertical line passing through the vertex.

4. Quadratic functions have exactly one increasing interval and exactly one decreasing interval.

5. The vertex is an extremum. If the parabola opens up, the function value at the vertex is a minimum value. If it opens down, the function value is a maximum value.

The general form of a quadratic function provides clues about the vertex, axis of symmetry, and extreme value but requires investigation in order to find them.

Thus, a potentially more useful form of a quadratic function might provide direct information about these properties.

Compare the graph of a basic squaring function.2f x x

To the graph of a transformed squaring function.22 4f x x


Vertex Form of a Quadratic Function (also called Standard Form):

2f x a x h k

If a function is written in vertex form, how can we find the properties of parabola?

1. Direction of Opening: If a > 0, the parabola opens up.

If a < 0, the parabola opens down.2. Vertex of the Parabola:

Vertex is at (h, k).3. Axis of Symmetry.

The axis of symmetry is x = h.4. Increasing/Decreasing:

If a > 0, f de , h] andincreases over [h, ). If a < 0. f , h] anddecreases over [h, ).

5. Extreme function value: If the parabola opens up, f(h) = k is theminimum value of f. If it opens down, f(h) = k is the maximum value of f.

By applying the completing the square technique, it is possible to rewrite a quadratic function into vertex form.

Steps to put a quadratic function into vertex form:

1. Group the x² and x terms.

2. As relevant, factor a from the group.

3a. Complete the square inside the parentheses.

3b. Compensate the function outside the parentheses.

4. Factor and simplify.


Ex. 1 Rewrite the function in vertex form and then identify the properties associated with the function.

2 6 7f x x x

Vertex: ( ___ , ___ ) Axis of Symmetry: ______

Increasing: _________ Decreasing: _________

Extremum is a __________ value of _________

Intercepts: ______________________________



22 8 10f x x x




Intercepts: ______________________________



23 6 2f x x x




Intercepts: ______________________________



213 4 7f x x x




Intercepts: ______________________________


When given the vertex of a quadratic function, it is possible to uniquely find the rule of the function if you know only one other point on the parabola.

Ex. 5 Find the function whose graph is a parabola with vertex 3,4 and passes through the point 1, 8 .

Ex. 6 Find the function whose graph is a parabola with vertex 1, 2 and passes through the point 2,4 .


Topic 29: Applications of Quadratic Functions

Many scenarios can be modeled using quadratic functions.

The trajectory of a launched object or the position of a dropped item often follow paths (over distance or time) modeled by a parabola.

For a launched object, it may be of interest where the maximum height occurs, which of course occurs at the vertex. For a quadratic function

2 ,f x ax bx c the vertex is at 2 2, .b ba af

Also potentially of interest is when the item reaches the ground. This occurs at the x-intercept of the quadratic function.

In economics, a basic formula for revenue is ( )R x x p x where x is the number of units being

sold and p(x) is the price of the item based on the number of units sold. Typical the price function is linear so the resulting revenue function is quadratic. A basic formula for cost is C x mx f where x is the number of units being produced, m is thevariable cost of producing those units, and f are the fixed costs. Although this model for cost is linear, because profit P is found by subtracting revenue minus costs (that is, P x R x C x ) many formulas for profit are also quadratic.

Finding a maximum value or when a function is zero may be of interest with these problems as well.


Ex. 1 An object is launched from a catapult with the height of the object, x seconds after launch, being estimated at

20.41 5.2 0.19f x x x meters.

Determine when the maximum height of the object is achieved and what the maximum height is. Then determine when the object hits the ground.


Ex. 2 A contractor for installing junction boxes determines that installing x boxes each day will result in profits of

20.2 28 480P x x x dollars.

Determine when the contractor breaks even each day (that is, when his profit is zero). Then determine what his maximum daily profit can be and how many boxes must be installed to produce that profit.


Ex. 3 The cost of producing t-shirts is $4 per shirt after fixed costs of $200. A seller determines that to sell x t-shirts, the price of each shirt can be found by p(x) = 60 – 0.5x.

Find the profit function for selling xshirts. Then assuming the seller wants to maximize profits, determine how many shirts at what price will make the most money, and how much she’ll make.


Topic 30: Solving Polynomial Equations

Some polynomial functions can by first finding a common factor and then applying other techniques as appropriate.

Ex. 1 Solve.

5 264x x

Ex. 2 Solve.

5 4 33 10 0x x x


The next two equations involve expressions that are factorable by the grouping method.

Ex. 3 Solve.

3 22 18 9 0x x x

Ex. 4 Solve.

3 22 3 6 0x x x


Topic 31: Solving Polynomial Inequalities

Polynomial inequalities, unlike the inequalities we have solved previously, must be solved in a very different manner than other inequalities. Whereasour approach to solving most inequalities is relatively straight-forward, our technique for solving nonlinear inequalities attacks the problem “sideways”, in a manner of speaking.

Process for solving polynomial inequalities:

1. As necessary, rewrite so that one side is zero.2. As necessary, factor the expression. 3. Set each factor equal to zero and solve to find

boundary values for the solution intervals. 4. Draw a number line and mark the boundaries

with an appropriate dot (closed if “equal to”, open if not).

5. Pick a test value from each interval formed by the boundaries and evaluate the expression using each one. Only the sign of the evaluation is relevant (all positive numbers are greater than zero, all negatives are less than zero.)

6. Identify the solution intervals.


Ex. 1 Solve. Present solutions graphically &in interval notation.

2 1 4 0x x


Ex. 2 Solve. Present solutions graphically & in interval notation.

3 22 4 8x x x


Ex. 3 Solve. Present solutions graphically & in interval notation.

3 25 14 0x x x

fx x x 3 x b xba x a x b - university of north georgia · standard form): xh y k r 222 form reveals...

Documents