Slide 8- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 1...

Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Quadratic Functions and Equations

11.1 Quadratic Equations

11.2 The Quadratic Formula

11.3 Studying Solutions of Quadratic Equations

11.4 Applications Involving Quadratic Equations

11.5 Equations Reducible to Quadratic

11.6 Quadratic Functions and Their Graphs

11.7 More About Graphing Quadratic Functions

11.8 Problem Solving and Quadratic Functions

11.9 Polynomial and Rational Inequalities

11


Quadratic Equations

The Principle of Square Roots

Completing the Square

Problem Solving

11.1

4Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The Principle of Square Roots

Let’s consider x2 = 25. We know that the number 25 has two real-number square roots, 5 and −5, the solutions of the equation. Thus we see that square roots can provide quick solutions for equations of the type x2 = k.


The Principle of Square RootsFor any real number k, if x2 = k, then

or .x k x k


Solution

Solve 5x 2 = 15. Give exact solutions and

approximations to three decimal places.

25 15x 2 3x

3 or 3.x x

We often use the symbol to represent both solutions.

3

The solutions are which round to 1.732 and –1.732. The check is left to the student.

3,

Isolating x2

Using the principle of square roots

Example


Solution

Solve 16x2 + 9 = 0.

216 9 0x 2 9

16x

9 9 or

16 16x x

The solutions are The check is left to the student.

3.

4i

3 3 or .

4 4x i x i Recall that 1 . i

Example


The Principle of Square Roots (Generalized Form)For any real number k and any algebraic expression X:

If X 2 = k, then or .X k X k


Solution

Solve: (x + 3)2 = 7.

2( 3) 7x

3 7 or 3 7x x

The solutions are 3 7.

3 7 or 3 7.x x

Example



Not all quadratic equations can be solved as in the previous examples. By using a method called completing the square, we can use the principle of square roots to solve any quadratic equation.


Solution

Solve x2 + 10x + 4 = 0.

2( 5) 21x

2 10 4 0x x 2 10 4x x

x2 + 10x + 25 = –4 + 25

5 21 or 5 21x x

5 21. The solutions are

Using the principle of square roots

Factoring

Adding 25 to both sides.

Example


To Solve a Quadratic Equation in x byCompleting the Square

1. Isolate the terms with variables on one side of the equation, and arrange them in descending order.

2. Divide both sides by the coefficient of x2 if that coefficient is not 1.

3. Complete the square by taking half of the coefficient of x and adding its square to both sides.


4. Express the trinomial as the square of a binomial (factor the trinomial) and simplify the other side.

5. Use the principle of square roots (find the square roots of both sides).

6. Solve for x by adding or subtracting on both sides.


Problem SolvingAfter one year, an amount of money P, invested at 4% per year, is worth 104% of P, or P(1.04). If that amount continues to earn 4% interest per year, after the second year the investment will be worth 104% of P(1.04), or P(1.04)2. This is called compounding interest since after the first period, interest is earned on both the initial investment and the interest from the first period. Generalizing, we have the following.


The Compound Interest Formula

If the amount of money P is invested at interest rate r, compounded annually, then in t years, it will grow to the amount A given by

(r is written in decimal notation.)(1 ) .tA P r


Solution

Jackson invested $5800 at an interest rate of r, compounded annually. In 2 yr, it grew to $6765. What was the interest rate?

1. Familiarize. We are already familiar with the compound-interest formula.

(1 )tA P r 6765 = 5800(1 + r)2

2. Translate. The translation consists of substituting into the formula:

Example


3. Carry out. Solve for r:

6765/5800 = (1 + r)2

6765/5800 1 r

1 6765/5800 r

.08 or 2.08r r

4. Check. Since the interest rate cannot negative, we need only to check .08 or 8%. If $5800 were invested at 8% compounded annually, then in 2 yr it would grow to 5800(1.08)2, or $6765. The number 8% checks.

5. State. The interest rate was 8%.


The Quadratic Formula

Solving Using the Quadratic Formula

Approximating Solutions

11.2


Solving Using the Quadratic Formula

Each time we solve by completing the square, the procedure is the same. When a procedure is repeated many times, a formula can often be developed to speed up our work.

If we begin with a quadratic equation in standard form, ax2 + bx + c = 0, and solve by completing the square we arrive at the quadratic formula.


The Quadratic FormulaThe solutions of ax2 + bx + c = 0,

are given by 2 4

.2

b b acx

a

0,a


Solution

Solve 3x2 + 5x = 2 using the quadratic formula.

First determine a, b, and c:

3x2 + 5x – 2 = 0;

a = 3, b = 5, and c = –2.

2 4

2

b b acx

a

25 5 34( )( 2)

2( )3

x Substituting

Example


5 25 24 5 49

6 6x

5 7

6x

5 7 5 7 or

6 6x x

2 12 or

6 6x x

1 or 2.

3x x

The solutions are 1/3 and –2. The check is left to the student.


To Solve a Quadratic Equation 1. If the equation can easily be written in the form

ax2 = p or (x + k)2 = d, use the principle of square roots.

2. If step (1) does not apply, write the equation in the form ax2 + bx + c = 0.

3. Try factoring using the principle of zero products.

4. If factoring seems difficult or impossible, use the quadratic formula. Completing the square can also be used.

The solutions of a quadratic equation can always be found using the quadratic formula. They cannot always be found by factoring.


Recall that a second-degree polynomial in one variable is said to be quadratic. Similarly, a second-degree polynomial function in one variable is said to be a quadratic function.


SolutionFirst determine a, b, and c:

x2 – 2x + 7 = 0;

a = 1, b = –2, and c = 7.

2 4

2

b b acx

a

22 ( 2) 4(1)(7)

2(1)x

Solve x2 + 7 = 2x using the quadratic formula.

Substituting

Example


2 4 28 2 24

2 2x

2 2 6

2

ix

2 2 6 1 6 .

2 2x i i

The solutions are

The check is left to the student.

1 6 and 1 6 .i i


Approximating Solutions

When the solution of an equation is irrational, a rational-number approximation is often useful. This is often the case in real-world applications similar to those found in section 8.4.


Solution

2 31.

3

Use a calculator to approximate

2 312.522588.

3

Take the time to familiarize yourself with your calculator:

Example


Studying Solutions of Quadratic Equations

The Discriminant

Writing Equations from Solutions

11.3


The Discriminant

It is sometimes enough to know what type of number a solution will be, without actually solving the equation. From the quadratic formula, b2 – 4ac, is known as the discriminant. The discriminant determines what type of number the solutions of a quadratic equation are. The cases are summarized on the next slide.


Discriminant b2 – 4ac

Nature of Solutions

0 One solution; a rational number

Positive Perfect square Not a perfect

square

Two different real-number solutions Solutions are rational Solutions are irrational conjugates

Negative Two different imaginary-number solutions (complex conjugates)


Solution

For the equation 4x2 – x + 1 = 0, determine what type of number the solutions are and how many solutions exist.

First determine a, b, and c: a = 4, b = –1, and c = 1. Compute the discriminant:

b2 – 4ac = (–1)2 – 4(4)(1) = –15.

Since the discriminant is negative, there are two imaginary-number solutions that are complex conjugates of each other.

Example


Solution

For the equation 5x2 – 10x + 5 = 0, determine what type of number the solutions are and how many solutions exist.

First determine a, b, and c: a = 5, b = –10, and c = 5. Compute the discriminant:

b2 – 4ac = (–10)2 – 4(5)(5) = 0.

There is exactly one solution, and it is rational. This indicates that 5x2 – 10x + 5 = 0 can be solved by factoring.

Example


Solution

For the equation 2x2 + 7x – 3 = 0, determine what type of number the solutions are and how many solutions exist.

First determine a, b, and c: a = 2, b = 7, and c = –3.

Compute the discriminant:

b2 – 4ac = (7)2 – 4(2)(–3) = 73.

The discriminant is a positive number that is not a perfect square. Thus there are two irrational solutions that are conjugates of each other.

Example


Writing Equations from Solutions

We know by the principle of zero products that (x – 1)(x + 4) = 0 has solutions 1 and -4. If we know the solutions of an equation, we can write an equation, using the principle in reverse.


Solution

Find an equation for which 5 and –4/3 are solutions.

x = 5 or x = –4/3

x – 5 = 0 or x + 4/3 = 0

(x – 5)(x + 4/3) = 0

x2 – 5x + 4/3x – 20/3 = 0

3x2 – 11x – 20 = 0

Get 0’s on one side

Using the principle of zero products

Multiplying

Combining like terms and clearing fractions

Example


Solution

Find an equation for which 3i and –3i are solutions.

x = 3i or x = –3i

x – 3i = 0 or x + 3i = 0

(x – 3i)(x + 3i) = 0

x2 – 3ix + 3ix – 9i2 = 0

x2 + 9 = 0

Get 0’s on one side

Using the principle of zero products

Multiplying

Combining like terms

Example


Applications Involving Quadratic Equations

Solving Problems

Solving Formulas

11.4


Solving Problems

As we found in Section 6.7, some problems translate to rational equations. The solution of such rational equations can involve quadratic equations.


Solution

Cade traveled 48 miles at a certain speed. If he had gone 4 mph faster, the trip would have taken 1 hr less. Find Cade’s average speed.

1. Familiarize. As in Section 6.5, we can create a table. Let r represent the rate, in miles per hour, and t the time, in hours for Cade’s trip.

Distance Speed Time

48 r t

48 r + 4 t – 1

48/r t48

41

rt

Example


2. Translate. From the table we obtain48

rt

and48

4 .1

rt

3. Carry out. A system of equations has been formed. We substitute for r from the first equation into the second and solve the resulting equation: 48 48

41t t

48 48( 1) 4 ( 1)

1t t t t

t t

Multiplying by the LCD


48 48( 1) ( 1) 4 ( 1)

1t t t t t t

t t

48( 1) 4 ( 1) 48t t t t

248 48 4 4 48t t t t 24 4 48 0t t

2 12 0t t

( 4)( 3) 0t t

4 0 or 3 0t t

4 or 3 t t


4. Check. Note that we solved for t, not r as required. Since negative time has no meaning here, we disregard the –3 and use 4 to find r:

48 48

4r

t 12 mph.

To see if 12 mph checks, we increase the speed 4 mph to 16 and see how long the trip would have taken at that speed:

483 hr.

16r

This is 1 hr less than the trip actually took, so the answer checks.

5. State. Cade traveled at 12 mph.


Solving Formulas

Recall that to solve a formula for a certain letter, we use the principles for solving equations to get that letter alone on one side.


Solution

Solve 215 for .

2D n n n

215

2D n n

22 5D n n 20 5 2 n n D

25 ( 5) 4(1)( 2 )

2(1)

Dn

5 25 8

2

Dn

Multiplying both sides by 2

Writing standard form

Using the quadratic formula

Example


To Solve a Formula for a Letter – Say, h1. Clear fractions and use the principle of powers, as

needed. Perform these steps until radicals containing h are gone and h is not in any denominator.

2. Combine all like terms.3. If the only power of h is h1, the equation can be

solved as in Sections 2.3 and 7.5. 4. If h2 appears but h does not, solve for h2 and use

the principle of square roots to solve for h.5. If there are terms containing both h and h2, put

the equation in standard form and use the quadratic formula.


Equations Reducible to Quadratic

Recognizing Equations in Quadratic Form

Radical Equations and Rational Equations

11.5


Recognizing Equations in Quadratic Form

Certain equations that are not really quadratic can be thought of in such a way that they can be solved as quadratic. For example, because the square of x2 is x4, the equation x4 – 5x2 + 4 = 0 is said to be “quadratic in x2”:


x4 – 5x2 + 4 = 0

(x2)2 – 5(x2) + 4 = 0

u2 – 5u + 4 = 0.

The last equation can be solved by factoring or by the quadratic formula. Then, remembering that u = x2, we can solve for x. Equations that can be solved like this are reducible to quadratic, or in quadratic form.


Solution

Solve x4 – 5x2 + 4 = 0.

u2 – 5u + 4 = 0

Let u = x2. Then we solve by substituting u for x2

and u2 for x4:

(u – 1)(u – 4) = 0

u = 1 or u = 4

u – 1 = 0 or u – 4 = 0

Factoring

Principle of zero products

Example


x2 = 1 or x2 = 41 or 2 x x

To check, note that for both x = 1 and x = –1, we have x2 = 1 and x4 = 1. Similarly, for both x = 2 and x = –2, we have x2 = 4 and x4 = 16. Thus instead of making four checks, we need make only two.

Check:x = 1: x = 2:

x4 – 5x2 + 4 = 0

(1) – 5(1) + 4 = 0

x4 – 5x2 + 4 = 0

(16) – 5(4) + 4 = 0

The solutions are 1, –1, 2, and –2.

TRUE TRUE

Replace u with x2


Caution! A common error on problems like the previous example is to solve for u but forget to solve for x. Remember to solve for the original variable!


Radical and Rational Equations

Sometimes rational equations, radical equations, or equations containing exponents that are fractions are reducible to quadratic. It is especially important that answers to these equations be checked in the original equation.


Solution

Solve 8 9 0.x x

u2 – 8u – 9 = 0

(u – 9)(u +1) = 0

u = 9 or u = –1

u – 9 = 0 or u + 1 = 0

xLet u = . Then we solve by substituting u for and u2 for x:

x

Example


81 or 1 x x

Check:x = 81: x = 1:

The solution is 81.

FALSE

TRUE

9 or 1x x

8 9 0x x 8 9 0x x

81 8 81 9 0 81 8(9) 9 0

1 8 1 9 0

1 8 9 0 81 72 9 0


Solution

Solve 2 14 2 0.t t

u2 + 4u – 2 = 0

Let u = t −1. Then we solve by substituting u for t −1 and u2 for t −2:

24 (4) 4(1)( 2)

2(1)u

4 16 82 6

2u

Example


1 12 6 or 2 6t t

1 12 6 or 2 6

tt

1 1 or t .

2 6 2 6t

The check is left to the student.


To Solve an Equation That is Reducibleto Quadratic 1. Look for two variable expressions in the

equation. One expressions should be the square of the other.

2. Write down any substitutions that you are making.

3. Remember to solve for the variable that is used in the original equation.

4. Check possible answers in the original equation.


Quadratic Functions and Their Graphs

The Graph of f (x) = ax2

The Graph of f (x) = a(x – h)2

The Graph of f (x) = a(x – h)2 + k

11.6


The Graph of f (x) = ax2

All quadratic functions have graphs similar to y = x2. Such curves are called parabolas. They are U-shaped and symmetric with respect to a vertical line known as the parabola’s axis of symmetry. For the graph of f (x) = x2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola.


Solutionx f(x) = x2 (x, f(x))

01–12–2

02288

(0, 0)(1, 2)

(–1, 2)(2, 8)

(–2, 8)

Graph: 2( ) 2 .f x xExample

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-2

-1

1

2

3

4

5

6

7

8

0

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-2

-1

1

2

3

4

5

6

7

8

0

(0,0)

(−1, 2) (1, 2)

(−2, 8) (2, 8)

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-2

-1

1

2

3

4

5

6

7

8

0


Graph: 2( ) 3 .f x x

Solution

x f(x) (x, f(x))

01

–12

–2

0–3–3

–12–12

(0, 0)(1, –3)(–1, –3)(2, –12)

(–2, –12)

Example

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-14

-12

-10

-8

-6

-4

-2

2

0

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-14

-12

-10

-8

-6

-4

-2

2

0

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-14

-12

-10

-8

-6

-4

-2

2

0


2( ) f x x

4

3

6

2

5

1

2( ) 4f x x

21( )

4f x x

2( )f x x


Graphing f (x) = ax2 The graph of f (x) = ax2 is a parabola with x = 0 as its axis of symmetry. Its vertex is the origin.

For a > 0, the parabola opens upward. For a < 0,the parabola opens downward.

If |a| is greater than 1, the parabola is narrower than y = x2.If |a| is between 0 and 1, the parabola is wider than y = x2.


The Graph of f (x) = a(x – h)2

We could next consider graphs of f (x) = ax2 + bx + c, where b and c are not both 0. It turns out to be convenient to first graph f (x) = a(x – h)2, where h is some constant. This allows us to observe similarities to the graphs drawn in previous slides.


Graph: 2( ) ( 2) .f x x

Solutionx f(x) = (x – 2)2

01

–1234

419014

vertex

Example

X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

1

2

3

4

5

6

7

8

9

10

0


Graphing f (x) = a(x – h)2

The graph of f (x) = a(x – h)2 has the same shapeas the graph of y = ax2.

• If h is positive, the graph of y = ax2 is shifted hunits to the right.

• If h is negative, the graph of y = ax2 is shifted |h| units to the left.

• The vertex is (h, 0) and the axis of symmetry is x = h.


The Graph of f (x) = a(x – h)2 + k

Given a graph of f (x) = a(x – h)2, what happens if we add a constant k? Suppose that we add 2. This increases f (x) by 2, so the curve is moved up. If k is negative, the curve is moved down. The axis of symmetry for the parabola remains x = h, but the vertex will be at (h, k), or equivalently (h, f (h)).


Graphing f (x) = a(x – h)2 + k

The graph of f (x) = a(x – h)2 + k has the same shapeas the graph of y = a(x – h)2.

• If k is positive, the graph of y = a(x – h)2 isshifted k units up.

• If k is negative, the graph of y = a(x – h)2 isshifted |k| units down.

• The vertex is (h, k), and the axis of symmetry is x = h.

• For a > 0, k is the minimum function value. For a < 0, the maximum function value is k.


Solution

Graph: and find the maximum function value.

21( ) ( 3) 1,

2 f x x

x

0–1–2–3–4–5

-11/2–3

–3/2–1

–3/2–3

vertex

Example

21( ) ( 3) 1

2f x x X

Y

-5 -4 -3 -2 -1 1 2 3 4 5

-8

-7

-6

-5

-4

-3

-2

-1

1

2

0

Maximum = −1


More About Graphing Quadratic Functions


Finding Intercepts

11.7



By completing the square, we can rewrite any polynomial ax2 + bx + c in the form a(x – h)2 + k. Once that has been done, the procedures discussed in Section 11.6 will enable us to graph any quadratic function.


Solution

Graph: 2( ) 2 1.f x x x

f (x) = x2 – 2x – 1

= (x2 – 2x) – 1

= (x2 – 2x + 1) – 1 – 1

= (x2 – 2x + 1 – 1) – 1

= (x – 1)2 – 2

The vertex is at (1, –2).

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

6

54

-4

-5

Example


Solution

Graph: 2( ) 2 6 3.f x x x

f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3

= –2(x2 – 3x + 9/4) – 3 + 18/4

= –2(x2 – 3x + 9/4 – 9/4) – 3

= –2(x – 3/2)2 + 3/2

The vertex is at (3/2, 3/2).

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

6

54

-4

-5

Example


The Vertex of a ParabolaThe vertex of a parabola given by f (x) = ax2 + bx + c is

• The x-coordinate of the vertex is –b/(2a).

• The axis of symmetry is x = -b/(2a).

• The second coordinate of the vertex is most commonly found by computing

24, or , .

2 2 2 4

b b b ac bf

a a a a

.2

bf

a


Finding Intercepts

For any function f, the y-intercept occurs at f (0). Thus for f (x) = ax2 + bx + c, the y-intercept is simply (0, c). To find x-intercepts, we look for points where y = 0 or f (x) = 0. Thus, for f (x) = ax2 + bx + c, the x-intercepts occur at those x-values for whichax2 + bx + c = 0


x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

6

54

-4

-5

f (x) = ax2 + bx + c

x - intercepts

y - intercept


Solution

Find the x- and y-intercepts of the graph of f (x) = x2 + 3x – 1.

The y-intercept is simply (0, f (0)), or (0, –1). To find the x-intercepts, we solve the equation: x2 + 3x – 1 = 0.

Since we are unable to solve by factoring, we use the quadratic formula to get 3 13 / 2.x

If graphing, we would approximate to get (–3.3, 0) and (0.3, 0).

Example


Problem Solving and Quadratic Functions

Maximum and Minimum Problems

Fitting Quadratic Functions to Data

11.8


Maximum and Minimum Problems

We have seen that for any quadratic function f, the value of f (x) at the vertex is either a maximum or a minimum. Thus problems in which a quantity must be maximized or minimized can be solved by finding the coordinates of the vertex, assuming the problem can be modeled with a quadratic function.


A farmer has 200 ft of fence with which to form a rectangular pen on his farm. If an existing fence forms one side of the rectangle, what dimensions will maximize the size of the area?

Example


Fitting Quadratic Functions to Data

Whenever a certain quadratic function fits a situation, that function can be determined if three inputs and their outputs are known. Each of the given ordered pairs is called a data point.


Use the data points (0, 10.4), (3, 16.8), and (6, 12.6) to find a quadratic function that fits the data.

Example

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