Chapter 8 Notes

CHAPTER 8 QUADRATIC FUNCTIONS AND EQUATIONS 8.1) Square Root Method & Completing the Square A Quadratic Equation is an equation in the form ax² + bx + c = 0. In this chapter we will look at different ways to solve these equations. They will include the following: 1) 2) 3) 4) 5)

Factoring Square Root Method Completing the Square Quadratic Formula Graphing

SOLUTIONS BY FACTORING: The first approach we try in solving a quadratic equation is by factoring. Try the following: Examples: 1) Solve for a: a ² + 8a + 7 = 0

2) Solve for x: 4 x ² + 14 x = −12

SOLUTIONS BY THE SQUARE ROOT METHOD: This method can be used for equations that are or can be put in the form ( )² = c The opposite of squaring an expression is to take the square root so we follow the following steps to solve equations in the form ( )² = c 1) Factor into a perfect square ( )² = c 2) Take the square root of both sides remembering the ± 3) Solve the resulting equation Consider the following: x² = 25

EXAMPLES: 1) Solve: (b − 3)² = 200

2) Solve: 3 x ² − 7 = 0

3) Solve: x ² − 2 x + 1 = −2

COMPLETING THE SQUARE: The goal for completing the square is to change any quadratic equation into a perfect square so it can be solved using the square root method. Procedure: Given the equation in the form ax ² + bx + c = 0 1) Move the constant c to the other side 2) The leading coefficient a must be 1. If not, divide each term by a 2 1   b 3) Add the value  2  to both sides 4) Factor the left side and solve the resulting equation using the square root method EXAMPLES: 1) Solve: x ² + 10 x + 9 = 0

2) Solve: 2 x ² + 12 x + 14 = 0

e3300f439b271000000000000000000000000000000000000000000000000000000000000000000000

3) Solve: 3 x ² + 5 x − 2 = 0

8.2) Quadratic Formula The Quadratic Formula is a shortcut for completing the square. The procedure of completing the square was performed on an arbitrary equation ax ² + bx + c = 0 to obtain the following form of all solutions to a quadratic equation:

EXAMPLES: 1) Solve: − 2 x ² − 4 x + 9 = 0

2) Solve: − 6 x ² = −13x − 5

3) Solve: 3 x ² + 5 x + 3 = 0

4. Write a quadratic equation with the given solution set: {-2, 6}

5. Write a quadratic equation with the solution set: {-8i, 8i}

6. Write the quadratic equation with the solution set: {2 + i, 2 – i}

8.3) Quadratic Functions and Their Graphs The graph of a quadratic function is a U-shaped curve called a parabola. In this section we will develop techniques for graphing such functions. A parabola has the following properties: 1) It is symmetric to the vertical line called the axis of symmetry 2) The minimum or maximum point is called the vertex Graphically we have:

f ( x) = ax ² Let us discover some other properties about the graph of the quadratic function. Using your graphing calculator, graph the following functions in the form f ( x ) = ax ² f ( x) = x ²

f ( x) = 2 x ²

f ( x) =

1 x² 2

f ( x ) = −2 x ²

From these graphs we can see the following: 1. The vertex of the parabola is ______________________________________________ 2. The axis of symmetry is__________________________________________________ 3. The parabola opens up when _____________________________________________ 4. The parabola opens down when ___________________________________________ 5. The coefficient of x² determines the ________________________________________ a) if | a | > 1 the parabola is______________________

f ( x) = a( x − h)²

b) if | a | < 1 the parabola is ______________________

Now let’s graph the following functions in the form f ( x) = a ( x − h)² along with the graph of f ( x) = x ² and make some observations: f ( x) = ( x − 3)²

f ( x) = ( x + 4)²

f ( x) = 2( x + 1)²

f ( x) = −( x − 2)²

From the graphs we should be able to conclude the following: 1. The vertex of the parabola is ______________________________________________ 2. The axis of symmetry is__________________________________________________ 3. If h is positive the graph is_________________________________________________ 4. If h is negative the graph is ________________________________________________

f ( x) = a( x − h)² + k Now let’s graph the following functions in the form f ( x) = a ( x − h)² + k along with the graph of the previous functions and make some observations: f ( x) = ( x − 3)² + 2

f ( x) = ( x + 4)² − 1

f ( x) = 2( x + 1)² + 3

From the graphs we should be able to conclude the following: 1. The vertex of the parabola is ______________________________________________ 2. The axis of symmetry is__________________________________________________ 3. If h is positive the graph is_________________________________________________ 4. If h is negative the graph is ________________________________________________ 5. If k is positive the graph is _________________________________________________ 6. If k is negative the graph is _________________________________________________ To conclude, we have discovered that for a quadratic equation in the form: f ( x ) = a ( x − h)² + k 1) a determines which way the parabola opens. If a is positive the parabola opens upward and if a is negative the parabola opens downward. As |a| increases in value the parabola is narrower and as |a| decreases the parabola is wider. 2) The vertex of the parabola is (h, k) 3) The axis of symmetry is x=h Examples: Graph the following by hand: 1) f ( x ) = 2( x − 1)² + 3

2)

f ( x) =

1 ( x + 1)² 2

Graphing Quadratic Functions in the form: ax² + bx + c = 0 By completing the square we can put any quadratic equation in the ax ² + bx + c = 0 into the form f ( x) = a ( x − h)² + k so we can graph the function using the properties we discussed in the last section. Examples: Put the following functions in the form a ( x − h)² + k by completing the square: 1)

f ( x ) = x ² − 10 x + 21

2) f ( x ) = 2 x ² − 16 x + 23

If we complete the square to the general form ax² + bx + c we can obtain a formula for the coordinates of the vertex. We will find theses to be the following:  − b 4ac − b²  ,   (h, k ) =  2a 4a  Since this is an ordered pair, if you find one of the values you can find the other by making a substitution. Examples: Find the vertex for the following functions: 1) f ( x ) = −3 x ² − 7 x + 2

2)

f ( x) =

1 x ² − 3x + 2 2

APPLICATIONS We have seen from the previous section that the vertex represents either a maximum or a minimum. If the parabola opens upward (a is positive) the vertex will be a minimum. If the parabola opens downward (a is negative) the vertex will be a maximum. Using this concept will vertices let’s solve the following problems. Examples: 1) An artist is designing a rectangular stained-glass window with a perimeter of 84 in. What dimensions will yield the maximum area?

2) A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the maximum area that the farmer can enclose with 40 ft of fence? What should the dimensions of the garden be in order to yield this area?

3) A person standing close to the edge on the top of a 200-foot building throws a baseball vertically upward. The quadratic function s(t) = -16t² + 64 t + 200 models the ball height above the ground, s(t), in feet, t seconds after it was thrown. a) After how many seconds does the ball reach its maximum height? What is the maximum height?

b) How many seconds does it take until the ball finally hits the ground?

c) Find s(0) and describe what it means.

d) Graph the quadratic function starting with t = 1 and ending when it hits the ground.

8.5) Polynomial and Rational Inequalities To solve quadratic inequalities, we are looking for intervals representing the solutions. The boundaries for these intervals are called critical values. Critical values are values that make the function zero or undefined. The following steps solve these inequalities: 1) Bring all terms to one side 2) Find the critical values (where the equation equals zero or is undefined) 3) Place the critical values on the values on the number line to distinguish possible intervals 4) Test each interval for solutions 5) Write the solution intervals in equality form

Examples:

1) Solve ( x + 3)( x − 5) > 0

2) Solve − 3 x( x + 1)( x − 4) > 0

3) Solve x ² − 4 x < 12

4) Solve:

1 >0 x+4

5 − 2x ≤0 4) Solve: 4 x + 3

5) Divers in Acapulco, Mexico, dive headfirst from the top of a cliff 87 feet above the Pacific Ocean. The function s(t) = -16t² + 8t + 87 models a divers height above the ocean, s(t), in feet, t seconds after leaping. During which time period will the diver’s height exceed that of the cliff?