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- Quiz: Solving Quadratic Equations
- Preliminaries
- Quiz: Preliminaries
- Properties of Basic Mathematical Operations
- Quiz: Properties of Basic Mathematical Operations
- Multiplying and Dividing Using Zero
- Quiz: Multiplying and Dividing Using Zero
- Powers and Exponents
- Quiz: Powers and Exponents
- Square Roots and Cube Roots
- Quiz: Square Roots and Cube Roots
- Grouping Symbols
- Quiz: Grouping Symbols
- Divisibility Rules
- Quiz: Divisibility Rules
- Signed Numbers (Positive Numbers and Negative Numbers)
- Quiz: Signed Numbers (Positive Numbers and Negative Numbers)
- Quiz: Fractions
- Simplifying Fractions and Complex Fractions
- Quiz: Simplifying Fractions and Complex Fractions
- Quiz: Decimals
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- Quiz: Scientific Notation
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- Quiz: Variables and Algebraic Expressions
- Evaluating Expressions
- Quiz: Evaluating Expressions
- Quiz: Equations
- Ratios and Proportions
- Quiz: Ratios and Proportions
- Solving Systems of Equations (Simultaneous Equations)
- Quiz: Solving Systems of Equations (Simultaneous Equations)
- Quiz: Monomials
- Polynomials
- Quiz: Polynomials
- Quiz: Factoring
- What Are Algebraic Fractions?
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- Quiz: Operations with Algebraic Fractions
- Inequalities
- Quiz: Inequalities
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- Quiz: Graphing on a Number Line
- Absolute Value
- Quiz: Absolute Value
- Solving Equations Containing Absolute Value
- Coordinate Graphs
- Quiz: Coordinate Graphs
- Linear Inequalities and Half-Planes
- Quiz: Linear Inequalities and Half-Planes
- Quiz: Functions
- Quiz: Variations
- Introduction to Roots and Radicals
- Simplifying Square Roots
- Quiz: Simplifying Square Roots
- Operations with Square Roots
- Quiz: Operations with Square Roots
- Solving Quadratic Equations
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- 9.3 Solve Quadratic Equations Using the Quadratic Formula
- Introduction
- 1.1 Use the Language of Algebra
- 1.2 Integers
- 1.3 Fractions
- 1.4 Decimals
- 1.5 Properties of Real Numbers
- Key Concepts
- Review Exercises
- Practice Test
- 2.1 Use a General Strategy to Solve Linear Equations
- 2.2 Use a Problem Solving Strategy
- 2.3 Solve a Formula for a Specific Variable
- 2.4 Solve Mixture and Uniform Motion Applications
- 2.5 Solve Linear Inequalities
- 2.6 Solve Compound Inequalities
- 2.7 Solve Absolute Value Inequalities
- 3.1 Graph Linear Equations in Two Variables
- 3.2 Slope of a Line
- 3.3 Find the Equation of a Line
- 3.4 Graph Linear Inequalities in Two Variables
- 3.5 Relations and Functions
- 3.6 Graphs of Functions
- 4.1 Solve Systems of Linear Equations with Two Variables
- 4.2 Solve Applications with Systems of Equations
- 4.3 Solve Mixture Applications with Systems of Equations
- 4.4 Solve Systems of Equations with Three Variables
- 4.5 Solve Systems of Equations Using Matrices
- 4.6 Solve Systems of Equations Using Determinants
- 4.7 Graphing Systems of Linear Inequalities
- 5.1 Add and Subtract Polynomials
- 5.2 Properties of Exponents and Scientific Notation
- 5.3 Multiply Polynomials
- 5.4 Dividing Polynomials
- Introduction to Factoring
- 6.1 Greatest Common Factor and Factor by Grouping
- 6.2 Factor Trinomials
- 6.3 Factor Special Products
- 6.4 General Strategy for Factoring Polynomials
- 6.5 Polynomial Equations
- 7.1 Multiply and Divide Rational Expressions
- 7.2 Add and Subtract Rational Expressions
- 7.3 Simplify Complex Rational Expressions
- 7.4 Solve Rational Equations
- 7.5 Solve Applications with Rational Equations
- 7.6 Solve Rational Inequalities
- 8.1 Simplify Expressions with Roots
- 8.2 Simplify Radical Expressions
- 8.3 Simplify Rational Exponents
- 8.4 Add, Subtract, and Multiply Radical Expressions
- 8.5 Divide Radical Expressions
- 8.6 Solve Radical Equations
- 8.7 Use Radicals in Functions
- 8.8 Use the Complex Number System
- 9.1 Solve Quadratic Equations Using the Square Root Property
- 9.2 Solve Quadratic Equations by Completing the Square
- 9.4 Solve Equations in Quadratic Form
- 9.5 Solve Applications of Quadratic Equations
- 9.6 Graph Quadratic Functions Using Properties
- 9.7 Graph Quadratic Functions Using Transformations
- 9.8 Solve Quadratic Inequalities
- 10.1 Finding Composite and Inverse Functions
- 10.2 Evaluate and Graph Exponential Functions
- 10.3 Evaluate and Graph Logarithmic Functions
- 10.4 Use the Properties of Logarithms
- 10.5 Solve Exponential and Logarithmic Equations
- 11.1 Distance and Midpoint Formulas; Circles
- 11.2 Parabolas
- 11.3 Ellipses
- 11.4 Hyperbolas
- 11.5 Solve Systems of Nonlinear Equations
- 12.1 Sequences
- 12.2 Arithmetic Sequences
- 12.3 Geometric Sequences and Series
- 12.4 Binomial Theorem

## Learning Objectives

By the end of this section, you will be able to:

- Solve quadratic equations using the Quadratic Formula
- Use the discriminant to predict the number and type of solutions of a quadratic equation
- Identify the most appropriate method to use to solve a quadratic equation

## Be Prepared 9.7

Before you get started, take this readiness quiz.

Evaluate b 2 − 4 a b b 2 − 4 a b when a = 3 a = 3 and b = −2 . b = −2 . If you missed this problem, review Example 1.21 .

## Be Prepared 9.8

Simplify: 108 . 108 . If you missed this problem, review Example 8.13 .

## Be Prepared 9.9

Simplify: 50 . 50 . If you missed this problem, review Example 8.76 .

Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for x.

We start with the standard form of a quadratic equation and solve it for x by completing the square.

- Quadratic Formula

The solutions to a quadratic equation of the form ax 2 + bx + c = 0, where a ≠ 0 a ≠ 0 are given by the formula:

To use the Quadratic Formula , we substitute the values of a , b , and c from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the formula is an equation. Make sure you use both sides of the equation.

## Example 9.21

How to solve a quadratic equation using the quadratic formula.

Solve by using the Quadratic Formula: 2 x 2 + 9 x − 5 = 0 . 2 x 2 + 9 x − 5 = 0 .

## Try It 9.41

Solve by using the Quadratic Formula: 3 y 2 − 5 y + 2 = 0 3 y 2 − 5 y + 2 = 0 .

## Try It 9.42

Solve by using the Quadratic Formula: 4 z 2 + 2 z − 6 = 0 4 z 2 + 2 z − 6 = 0 .

## Solve a quadratic equation using the quadratic formula.

- Step 1. Write the quadratic equation in standard form, ax 2 + bx + c = 0. Identify the values of a , b , and c .
- Step 2. Write the Quadratic Formula. Then substitute in the values of a , b , and c .
- Step 3. Simplify.
- Step 4. Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “ x =”.

## Example 9.22

Solve by using the Quadratic Formula: x 2 − 6 x = −5 . x 2 − 6 x = −5 .

## Try It 9.43

Solve by using the Quadratic Formula: a 2 − 2 a = 15 a 2 − 2 a = 15 .

## Try It 9.44

Solve by using the Quadratic Formula: b 2 + 24 = −10 b b 2 + 24 = −10 b .

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula . If we get a radical as a solution, the final answer must have the radical in its simplified form.

## Example 9.23

Solve by using the Quadratic Formula: 2 x 2 + 10 x + 11 = 0 . 2 x 2 + 10 x + 11 = 0 .

## Try It 9.45

Solve by using the Quadratic Formula: 3 m 2 + 12 m + 7 = 0 3 m 2 + 12 m + 7 = 0 .

## Try It 9.46

Solve by using the Quadratic Formula: 5 n 2 + 4 n − 4 = 0 5 n 2 + 4 n − 4 = 0 .

When we substitute a , b , and c into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

## Example 9.24

Solve by using the Quadratic Formula: 3 p 2 + 2 p + 9 = 0 . 3 p 2 + 2 p + 9 = 0 .

## Try It 9.47

Solve by using the Quadratic Formula: 4 a 2 − 2 a + 8 = 0 4 a 2 − 2 a + 8 = 0 .

## Try It 9.48

Solve by using the Quadratic Formula: 5 b 2 + 2 b + 4 = 0 5 b 2 + 2 b + 4 = 0 .

Remember, to use the Quadratic Formula, the equation must be written in standard form, ax 2 + bx + c = 0. Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

## Example 9.25

Solve by using the Quadratic Formula: x ( x + 6 ) + 4 = 0 . x ( x + 6 ) + 4 = 0 .

Our first step is to get the equation in standard form.

## Try It 9.49

Solve by using the Quadratic Formula: x ( x + 2 ) − 5 = 0 . x ( x + 2 ) − 5 = 0 .

## Try It 9.50

Solve by using the Quadratic Formula: 3 y ( y − 2 ) − 3 = 0 . 3 y ( y − 2 ) − 3 = 0 .

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

## Example 9.26

Solve by using the Quadratic Formula: 1 2 u 2 + 2 3 u = 1 3 . 1 2 u 2 + 2 3 u = 1 3 .

Our first step is to clear the fractions.

## Try It 9.51

Solve by using the Quadratic Formula: 1 4 c 2 − 1 3 c = 1 12 1 4 c 2 − 1 3 c = 1 12 .

## Try It 9.52

Solve by using the Quadratic Formula: 1 9 d 2 − 1 2 d = − 1 3 1 9 d 2 − 1 2 d = − 1 3 .

Think about the equation ( x − 3) 2 = 0. We know from the Zero Product Property that this equation has only one solution, x = 3.

We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to 0 gives just one solution. Notice that once the radicand is simplified it becomes 0 , which leads to only one solution.

## Example 9.27

Solve by using the Quadratic Formula: 4 x 2 − 20 x = −25 . 4 x 2 − 20 x = −25 .

Did you recognize that 4 x 2 − 20 x + 25 is a perfect square trinomial. It is equivalent to (2 x − 5) 2 ? If you solve 4 x 2 − 20 x + 25 = 0 by factoring and then using the Square Root Property, do you get the same result?

## Try It 9.53

Solve by using the Quadratic Formula: r 2 + 10 r + 25 = 0 . r 2 + 10 r + 25 = 0 .

## Try It 9.54

Solve by using the Quadratic Formula: 25 t 2 − 40 t = −16 . 25 t 2 − 40 t = −16 .

## Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant .

## Discriminant

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

## Using the Discriminant, b 2 − 4 ac , to Determine the Number and Type of Solutions of a Quadratic Equation

For a quadratic equation of the form ax 2 + bx + c = 0, a ≠ 0 , a ≠ 0 ,

- If b 2 − 4 ac > 0, the equation has 2 real solutions.
- if b 2 − 4 ac = 0, the equation has 1 real solution.
- if b 2 − 4 ac < 0, the equation has 2 complex solutions.

## Example 9.28

Determine the number of solutions to each quadratic equation.

ⓐ 3 x 2 + 7 x − 9 = 0 3 x 2 + 7 x − 9 = 0 ⓑ 5 n 2 + n + 4 = 0 5 n 2 + n + 4 = 0 ⓒ 9 y 2 − 6 y + 1 = 0 . 9 y 2 − 6 y + 1 = 0 .

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

Since the discriminant is positive, there are 2 real solutions to the equation.

Since the discriminant is negative, there are 2 complex solutions to the equation.

Since the discriminant is 0, there is 1 real solution to the equation.

## Try It 9.55

Determine the numberand type of solutions to each quadratic equation.

ⓐ 8 m 2 − 3 m + 6 = 0 8 m 2 − 3 m + 6 = 0 ⓑ 5 z 2 + 6 z − 2 = 0 5 z 2 + 6 z − 2 = 0 ⓒ 9 w 2 + 24 w + 16 = 0 . 9 w 2 + 24 w + 16 = 0 .

## Try It 9.56

Determine the number and type of solutions to each quadratic equation.

ⓐ b 2 + 7 b − 13 = 0 b 2 + 7 b − 13 = 0 ⓑ 5 a 2 − 6 a + 10 = 0 5 a 2 − 6 a + 10 = 0 ⓒ 4 r 2 − 20 r + 25 = 0 . 4 r 2 − 20 r + 25 = 0 .

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

## Methods for Solving Quadratic Equations

- Square Root Property
- Completing the Square

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is a x 2 = k a x 2 = k or a ( x − h ) 2 = k a ( x − h ) 2 = k we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

## Identify the most appropriate method to solve a quadratic equation.

- Step 1. Try Factoring first. If the quadratic factors easily, this method is very quick.
- Step 2. Try the Square Root Property next. If the equation fits the form a x 2 = k a x 2 = k or a ( x − h ) 2 = k , a ( x − h ) 2 = k , it can easily be solved by using the Square Root Property.
- Step 3. Use the Quadratic Formula . Any other quadratic equation is best solved by using the Quadratic Formula.

The next example uses this strategy to decide how to solve each quadratic equation.

## Example 9.29

Identify the most appropriate method to use to solve each quadratic equation.

ⓐ 5 z 2 = 17 5 z 2 = 17 ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0 ⓒ 8 u 2 + 6 u = 11 . 8 u 2 + 6 u = 11 .

ⓐ 5 z 2 = 17 5 z 2 = 17

Since the equation is in the a x 2 = k , a x 2 = k , the most appropriate method is to use the Square Root Property.

ⓑ 4 x 2 − 12 x + 9 = 0 4 x 2 − 12 x + 9 = 0

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

## Try It 9.57

ⓐ x 2 + 6 x + 8 = 0 x 2 + 6 x + 8 = 0 ⓑ ( n − 3 ) 2 = 16 ( n − 3 ) 2 = 16 ⓒ 5 p 2 − 6 p = 9 . 5 p 2 − 6 p = 9 .

## Try It 9.58

ⓐ 8 a 2 + 3 a − 9 = 0 8 a 2 + 3 a − 9 = 0 ⓑ 4 b 2 + 4 b + 1 = 0 4 b 2 + 4 b + 1 = 0 ⓒ 5 c 2 = 125 . 5 c 2 = 125 .

Access these online resources for additional instruction and practice with using the Quadratic Formula.

- Using the Quadratic Formula
- Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions
- Discriminant in Quadratic Formula

## Practice Makes Perfect

In the following exercises, solve by using the Quadratic Formula.

4 m 2 + m − 3 = 0 4 m 2 + m − 3 = 0

4 n 2 − 9 n + 5 = 0 4 n 2 − 9 n + 5 = 0

2 p 2 − 7 p + 3 = 0 2 p 2 − 7 p + 3 = 0

3 q 2 + 8 q − 3 = 0 3 q 2 + 8 q − 3 = 0

p 2 + 7 p + 12 = 0 p 2 + 7 p + 12 = 0

q 2 + 3 q − 18 = 0 q 2 + 3 q − 18 = 0

r 2 − 8 r = 33 r 2 − 8 r = 33

t 2 + 13 t = −40 t 2 + 13 t = −40

3 u 2 + 7 u − 2 = 0 3 u 2 + 7 u − 2 = 0

2 p 2 + 8 p + 5 = 0 2 p 2 + 8 p + 5 = 0

2 a 2 − 6 a + 3 = 0 2 a 2 − 6 a + 3 = 0

5 b 2 + 2 b − 4 = 0 5 b 2 + 2 b − 4 = 0

x 2 + 8 x − 4 = 0 x 2 + 8 x − 4 = 0

y 2 + 4 y − 4 = 0 y 2 + 4 y − 4 = 0

3 y 2 + 5 y − 2 = 0 3 y 2 + 5 y − 2 = 0

6 x 2 + 2 x − 20 = 0 6 x 2 + 2 x − 20 = 0

2 x 2 + 3 x + 3 = 0 2 x 2 + 3 x + 3 = 0

2 x 2 − x + 1 = 0 2 x 2 − x + 1 = 0

8 x 2 − 6 x + 2 = 0 8 x 2 − 6 x + 2 = 0

8 x 2 − 4 x + 1 = 0 8 x 2 − 4 x + 1 = 0

( v + 1 ) ( v − 5 ) − 4 = 0 ( v + 1 ) ( v − 5 ) − 4 = 0

( x + 1 ) ( x − 3 ) = 2 ( x + 1 ) ( x − 3 ) = 2

( y + 4 ) ( y − 7 ) = 18 ( y + 4 ) ( y − 7 ) = 18

( x + 2 ) ( x + 6 ) = 21 ( x + 2 ) ( x + 6 ) = 21

1 3 m 2 + 1 12 m = 1 4 1 3 m 2 + 1 12 m = 1 4

1 3 n 2 + n = − 1 2 1 3 n 2 + n = − 1 2

3 4 b 2 + 1 2 b = 3 8 3 4 b 2 + 1 2 b = 3 8

1 9 c 2 + 2 3 c = 3 1 9 c 2 + 2 3 c = 3

16 c 2 + 24 c + 9 = 0 16 c 2 + 24 c + 9 = 0

25 d 2 − 60 d + 36 = 0 25 d 2 − 60 d + 36 = 0

25 q 2 + 30 q + 9 = 0 25 q 2 + 30 q + 9 = 0

16 y 2 + 8 y + 1 = 0 16 y 2 + 8 y + 1 = 0

Use the Discriminant to Predict the Number of Real Solutions of a Quadratic Equation

In the following exercises, determine the number of real solutions for each quadratic equation.

ⓐ 4 x 2 − 5 x + 16 = 0 4 x 2 − 5 x + 16 = 0 ⓑ 36 y 2 + 36 y + 9 = 0 36 y 2 + 36 y + 9 = 0 ⓒ 6 m 2 + 3 m − 5 = 0 6 m 2 + 3 m − 5 = 0

ⓐ 9 v 2 − 15 v + 25 = 0 9 v 2 − 15 v + 25 = 0 ⓑ 100 w 2 + 60 w + 9 = 0 100 w 2 + 60 w + 9 = 0 ⓒ 5 c 2 + 7 c − 10 = 0 5 c 2 + 7 c − 10 = 0

ⓐ r 2 + 12 r + 36 = 0 r 2 + 12 r + 36 = 0 ⓑ 8 t 2 − 11 t + 5 = 0 8 t 2 − 11 t + 5 = 0 ⓒ 3 v 2 − 5 v − 1 = 0 3 v 2 − 5 v − 1 = 0

ⓐ 25 p 2 + 10 p + 1 = 0 25 p 2 + 10 p + 1 = 0 ⓑ 7 q 2 − 3 q − 6 = 0 7 q 2 − 3 q − 6 = 0 ⓒ 7 y 2 + 2 y + 8 = 0 7 y 2 + 2 y + 8 = 0

In the following exercises, identify the most appropriate method (Factoring, Square Root, or Quadratic Formula) to use to solve each quadratic equation. Do not solve.

ⓐ x 2 − 5 x − 24 = 0 x 2 − 5 x − 24 = 0 ⓑ ( y + 5 ) 2 = 12 ( y + 5 ) 2 = 12 ⓒ 14 m 2 + 3 m = 11 14 m 2 + 3 m = 11

ⓐ ( 8 v + 3 ) 2 = 81 ( 8 v + 3 ) 2 = 81 ⓑ w 2 − 9 w − 22 = 0 w 2 − 9 w − 22 = 0 ⓒ 4 n 2 − 10 n = 6 4 n 2 − 10 n = 6

ⓐ 6 a 2 + 14 a = 20 6 a 2 + 14 a = 20 ⓑ ( x − 1 4 ) 2 = 5 16 ( x − 1 4 ) 2 = 5 16 ⓒ y 2 − 2 y = 8 y 2 − 2 y = 8

ⓐ 8 b 2 + 15 b = 4 8 b 2 + 15 b = 4 ⓑ 5 9 v 2 − 2 3 v = 1 5 9 v 2 − 2 3 v = 1 ⓒ ( w + 4 3 ) 2 = 2 9 ( w + 4 3 ) 2 = 2 9

## Writing Exercises

Solve the equation x 2 + 10 x = 120 x 2 + 10 x = 120

ⓐ by completing the square

ⓑ using the Quadratic Formula

ⓒ Which method do you prefer? Why?

Solve the equation 12 y 2 + 23 y = 24 12 y 2 + 23 y = 24

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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## 7.7: Modeling with Quadratic Functions

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## Learning Objectives

- Recognize characteristics of parabolas
- Understand how the graph of a parabola is related to its quadratic function
- Solve problems involving a quadratic function’s minimum or maximum value

in order to apply mathematical modeling to solve real-world applications.

Curved antennas, such as the ones shown in Figure \(\PageIndex{1}\), are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.

In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.

## Recognizing Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry . These features are illustrated in Figure \(\PageIndex{2}\).

The y-intercept is the point at which the parabola crosses the \(y\)-axis. The x-intercepts are the points at which the parabola crosses the \(x\)-axis. If they exist, the x-intercepts represent the zeros , or roots , of the quadratic function, the values of \(x\) at which \(y=0\).

## Example \(\PageIndex{1}\): Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown in Figure \(\PageIndex{3}\).

The vertex is the turning point of the graph. We can see that the vertex is at \((3,1)\). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is \(x=3\). This parabola does not cross the x-axis, so it has no zeros. It crosses the \(y\)-axis at \((0,7)\) so this is the y-intercept.

## Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions

The general form of a quadratic function presents the function in the form

\[f(x)=ax^2+bx+c\]

where \(a\), \(b\), and \(c\) are real numbers and \(a{\neq}0\). If \(a>0\), the parabola opens upward. If \(a<0\), the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.

The axis of symmetry is defined by \(x=−\frac{b}{2a}\). If we use the quadratic formula, \(x=\frac{−b{\pm}\sqrt{b^2−4ac}}{2a}\), to solve \(ax^2+bx+c=0\) for the x-intercepts, or zeros, we find the value of \(x\) halfway between them is always \(x=−\frac{b}{2a}\), the equation for the axis of symmetry.

Figure \(\PageIndex{4}\) represents the graph of the quadratic function written in general form as \(y=x^2+4x+3\). In this form, \(a=1\), \(b=4\), and \(c=3\). Because \(a>0\), the parabola opens upward. The axis of symmetry is \(x=−\frac{4}{2(1)}=−2\). This also makes sense because we can see from the graph that the vertical line \(x=−2\) divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, \((−2,−1)\). The x-intercepts, those points where the parabola crosses the x-axis, occur at \((−3,0)\) and \((−1,0)\).

The standard form of a quadratic function presents the function in the form

\[f(x)=a(x−h)^2+k\]

where \((h, k)\) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function .

As with the general form, if \(a>0\), the parabola opens upward and the vertex is a minimum. If \(a<0\), the parabola opens downward, and the vertex is a maximum. Figure \(\PageIndex{5}\) represents the graph of the quadratic function written in standard form as \(y=−3(x+2)^2+4\). Since \(x–h=x+2\) in this example, \(h=–2\). In this form, \(a=−3\), \(h=−2\), and \(k=4\). Because \(a<0\), the parabola opens downward. The vertex is at \((−2, 4)\).

The standard form is useful for determining how the graph is transformed from the graph of \(y=x^2\). Figure \(\PageIndex{6}\) is the graph of this basic function.

If \(k>0\), the graph shifts upward, whereas if \(k<0\), the graph shifts downward. In Figure \(\PageIndex{5}\), \(k>0\), so the graph is shifted 4 units upward. If \(h>0\), the graph shifts toward the right and if \(h<0\), the graph shifts to the left. In Figure \(\PageIndex{5}\), \(h<0\), so the graph is shifted 2 units to the left. The magnitude of \(a\) indicates the stretch of the graph. If \(|a|>1\), the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if \(|a|<1\), the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure \(\PageIndex{5}\), \(|a|>1\), so the graph becomes narrower.

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

\[\begin{align*} a(x−h)^2+k &= ax^2+bx+c \\[4pt] ax^2−2ahx+(ah^2+k)&=ax^2+bx+c \end{align*} \]

For the linear terms to be equal, the coefficients must be equal.

\[–2ah=b \text{, so } h=−\dfrac{b}{2a}. \nonumber\]

This is the axis of symmetry we defined earlier. Setting the constant terms equal:

\[\begin{align*} ah^2+k&=c \\ k&=c−ah^2 \\ &=c−a−\Big(\dfrac{b}{2a}\Big)^2 \\ &=c−\dfrac{b^2}{4a} \end{align*}\]

In practice, though, it is usually easier to remember that \(k\) is the output value of the function when the input is \(h\), so \(f(h)=k\).

## Definitions: Forms of Quadratic Functions

A quadratic function is a function of degree two. The graph of a quadratic function is a parabola.

- The general form of a quadratic function is \(f(x)=ax^2+bx+c\) where \(a\), \(b\), and \(c\) are real numbers and \(a{\neq}0\).
- The standard form of a quadratic function is \(f(x)=a(x−h)^2+k\).
- The vertex \((h,k)\) is located at \[h=–\dfrac{b}{2a},\;k=f(h)=f(\dfrac{−b}{2a}).\]

## HOWTO: Write a quadratic function in a general form

Given a graph of a quadratic function, write the equation of the function in general form.

- Identify the horizontal shift of the parabola; this value is \(h\). Identify the vertical shift of the parabola; this value is \(k\).
- Substitute the values of the horizontal and vertical shift for \(h\) and \(k\). in the function \(f(x)=a(x–h)^2+k\).
- Substitute the values of any point, other than the vertex, on the graph of the parabola for \(x\) and \(f(x)\).
- Solve for the stretch factor, \(|a|\).
- If the parabola opens up, \(a>0\). If the parabola opens down, \(a<0\) since this means the graph was reflected about the x-axis.
- Expand and simplify to write in general form.

## Example \(\PageIndex{2}\): Writing the Equation of a Quadratic Function from the Graph

Write an equation for the quadratic function \(g\) in Figure \(\PageIndex{7}\) as a transformation of \(f(x)=x^2\), and then expand the formula, and simplify terms to write the equation in general form.

We can see the graph of \(g\) is the graph of \(f(x)=x^2\) shifted to the left 2 and down 3, giving a formula in the form \(g(x)=a(x+2)^2–3\).

Substituting the coordinates of a point on the curve, such as \((0,−1)\), we can solve for the stretch factor.

\[\begin{align} −1&=a(0+2)^2−3 \\ 2&=4a \\ a&=\dfrac{1}{2} \end{align}\]

In standard form, the algebraic model for this graph is \(g(x)=\dfrac{1}{2}(x+2)^2–3\).

To write this in general polynomial form, we can expand the formula and simplify terms.

\[\begin{align} g(x)&=\dfrac{1}{2}(x+2)^2−3 \\ &=\dfrac{1}{2}(x+2)(x+2)−3 \\ &=\dfrac{1}{2}(x^2+4x+4)−3 \\ &=\dfrac{1}{2}x^2+2x+2−3 \\ &=\dfrac{1}{2}x^2+2x−1 \end{align}\]

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

We can check our work using the table feature on a graphing utility. First enter \(\mathrm{Y1=\dfrac{1}{2}(x+2)^2−3}\). Next, select \(\mathrm{TBLSET}\), then use \(\mathrm{TblStart=–6}\) and \(\mathrm{ΔTbl = 2}\), and select \(\mathrm{TABLE}\). See Table \(\PageIndex{1}\)

The ordered pairs in the table correspond to points on the graph.

## Exercise \(\PageIndex{2}\)

A coordinate grid has been superimposed over the quadratic path of a basketball in Figure \(\PageIndex{8}\). Find an equation for the path of the ball. Does the shooter make the basket?

Figure \(\PageIndex{8}\): Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes. (credit: modification of work by Dan Meyer)

The path passes through the origin and has vertex at \((−4, 7)\), so \(h(x)=–\frac{7}{16}(x+4)^2+7\). To make the shot, \(h(−7.5)\) would need to be about 4 but \(h(–7.5){\approx}1.64\); he doesn’t make it.

- Identify \(a\), \(b\), and \(c\).
- Find \(h\), the x-coordinate of the vertex, by substituting \(a\) and \(b\) into \(h=–\frac{b}{2a}\).
- Find \(k\), the y-coordinate of the vertex, by evaluating \(k=f(h)=f\Big(−\frac{b}{2a}\Big)\).

## Example \(\PageIndex{3}\): Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function \(f(x)=2x^2–6x+7\). Rewrite the quadratic in standard form (vertex form).

The horizontal coordinate of the vertex will be at

\[\begin{align} h&=–\dfrac{b}{2a} \\ &=-\dfrac{-6}{2(2)} \\ &=\dfrac{6}{4} \\ &=\dfrac{3}{2}\end{align}\]

The vertical coordinate of the vertex will be at

\[\begin{align} k&=f(h) \\ &=f\Big(\dfrac{3}{2}\Big) \\ &=2\Big(\dfrac{3}{2}\Big)^2−6\Big(\dfrac{3}{2}\Big)+7 \\ &=\dfrac{5}{2} \end{align}\]

Rewriting into standard form, the stretch factor will be the same as the \(a\) in the original quadratic.

\[f(x)=ax^2+bx+c \\ f(x)=2x^2−6x+7\]

Using the vertex to determine the shifts,

\[f(x)=2\Big(x–\dfrac{3}{2}\Big)^2+\dfrac{5}{2}\]

One reason we may want to identify the vertex of the parabola is that this point will inform us what the maximum or minimum value of the function is, \((k)\),and where it occurs, \((h)\).

## Exercise \(\PageIndex{3}\)

Given the equation \(g(x)=13+x^2−6x\), write the equation in general form and then in standard form.

\(g(x)=x^2−6x+13\) in general form; \(g(x)=(x−3)^2+4\) in standard form.

## Determining the Maximum and Minimum Values of Quadratic Functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola . We can see the maximum and minimum values in Figure \(\PageIndex{9}\).

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

## Example \(\PageIndex{5}\): Finding the Maximum Value of a Quadratic Function

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.

- Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length \(L\).
- What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram such as Figure \(\PageIndex{10}\) to record the given information. It is also helpful to introduce a temporary variable, \(W\), to represent the width of the garden and the length of the fence section parallel to the backyard fence.

a. We know we have only 80 feet of fence available, and \(L+W+L=80\), or more simply, \(2L+W=80\). This allows us to represent the width, \(W\), in terms of \(L\).

\[W=80−2L\]

Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so

\[\begin{align} A&=LW=L(80−2L) \\ A(L)&=80L−2L^2 \end{align}\]

This formula represents the area of the fence in terms of the variable length \(L\). The function, written in general form, is

\[A(L)=−2L^2+80L\].

The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since \(a\) is the coefficient of the squared term, \(a=−2\), \(b=80\), and \(c=0\).

To find the vertex:

\[\begin{align} h& =−\dfrac{80}{2(−2)} &k&=A(20) \\ &=20 & \text{and} \;\;\;\; &=80(20)−2(20)^2 \\ &&&=800 \end{align}\]

The maximum value of the function is an area of 800 square feet, which occurs when \(L=20\) feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.

This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure \(\PageIndex{11}\).

- Write a quadratic equation for revenue.
- Find the vertex of the quadratic equation.
- Determine the y-value of the vertex.

## Example \(\PageIndex{6}\): Finding Maximum Revenue

The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, \(p\) for price per subscription and \(Q\) for quantity, giving us the equation \(\text{Revenue}=pQ\).

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently \(p=30\) and \(Q=84,000\). We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, \(p=32\) and \(Q=79,000\). From this we can find a linear equation relating the two quantities. The slope will be

\[\begin{align} m&=\dfrac{79,000−84,000}{32−30} \\ &=−\dfrac{5,000}{2} \\ &=−2,500 \end{align}\]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

\[\begin{align} Q&=−2500p+b &\text{Substitute in the point $Q=84,000$ and $p=30$} \\ 84,000&=−2500(30)+b &\text{Solve for $b$} \\ b&=159,000 \end{align}\]

This gives us the linear equation \(Q=−2,500p+159,000\) relating cost and subscribers. We now return to our revenue equation.

\[\begin{align} \text{Revenue}&=pQ \\ \text{Revenue}&=p(−2,500p+159,000) \\ \text{Revenue}&=−2,500p^2+159,000p \end{align}\]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

\[\begin{align} h&=−\dfrac{159,000}{2(−2,500)} \\ &=31.8 \end{align}\]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

\[\begin{align} \text{maximum revenue}&=−2,500(31.8)^2+159,000(31.8) \\ &=2,528,100 \end{align}\]

This could also be solved by graphing the quadratic as in Figure \(\PageIndex{12}\). We can see the maximum revenue on a graph of the quadratic function.

## Example \(\PageIndex{10}\): Applying the Vertex and x-Intercepts of a Parabola

A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation \(H(t)=−16t^2+80t+40\).

When does the ball reach the maximum height? What is the maximum height of the ball? When does the ball hit the ground?

The ball reaches the maximum height at the vertex of the parabola. \[\begin{align} h &= −\dfrac{80}{2(−16)} \\ &=\dfrac{80}{32} \\ &=\dfrac{5}{2} \\ & =2.5 \end{align}\]

The ball reaches a maximum height after 2.5 seconds.

To find the maximum height, find the y-coordinate of the vertex of the parabola. \[\begin{align} k &=H(−\dfrac{b}{2a}) \\ &=H(2.5) \\ &=−16(2.5)^2+80(2.5)+40 \\ &=140 \end{align}\]

The ball reaches a maximum height of 140 feet.

To find when the ball hits the ground, we need to determine when the height is zero, \(H(t)=0\).

We use the quadratic formula.

\[\begin{align} t & =\dfrac{−80±\sqrt{80^2−4(−16)(40)}}{2(−16)} \\ & = \dfrac{−80±\sqrt{8960}}{−32} \end{align} \]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

\[t=\dfrac{−80-\sqrt{8960}}{−32} ≈5.458 \text{ or }t=\dfrac{−80+\sqrt{8960}}{−32} ≈−0.458 \]

The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure \(\PageIndex{16}\).

## Example \(\PageIndex{11}\): Using Technology to Find the Best Fit Quadratic Model

A ball is thrown into the air, and the following data is collected where x represents the time in seconds after the ball is thrown up and y represents the height in meters of the ball. We can use desmos to create a quadratic model that fits the given data.

On desmos, type the data into a table with the x-values in the first column and the y-values in the second column. Then, to tell desmos to compute a quadratic model, type in y 1 ~ a x 1 2 + b x 1 + c. You will get a result that looks like this:

You can go to this problem in desmos by clicking https://www.desmos.com/calculator/u8ytorpnhk . Looking at the results, the quadratic model that fits the data is \[y = -4.9 x^2 + 20 x + 1.5\]

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