Chapter 6

Exploring Quadratic Functions and Inequalities

6-1 Quadratic Functions

• Functions with the form y=ax2+bx+c are called quadratic functions and their graphs have a parabolic shape

• When we solve ax2+bx+c=0 we look for values of x that are x-intercepts (because we have y=0)

• The x-intercepts are called the solutions or roots of a quadratic equation

• A quadratic equation can have two real solutions, one real solution, or no real solutions

6-1 Quadratic Functions (cont.)

• On the calculator find roots using the ROOT menu– Choose a point to the left of the x-intercept and a point to

the right of the x-intercept to give a range in which the calculator will find the x-intercept

– Do this for each root you see on the graph

6-2 Example

Graph y= -x2 - 2x + 8 and find its roots.

Vertex: (-1, 9)Roots: (-4, 0) (2, 0)Viewing window:Xmin= -10Xmax=10Ymin= -10Ymax= 10

6-2 Problems

1. Find what size viewing window is needed to view y= x2 + 4x -15. Find the roots.

Window: Xmin= -10 Xmax= 10 Ymin= -20 Ymax= 10 Roots: -6.3589 and 2.3589

6-2 Solving Quadratic Equations by Graphing

• In a quadratic equation y=ax2+bx+c, ax2 is the quadratic term, bx is the linear term, and c is the constant term

• The axis of symmetry is a line that divides a parabola into two equal parts that would match exactly if folded over on each other

• The vertex is where the axis of symmetry meets the parabola

• The roots or zeros (or solutions) are found by solving the quadratic equation for y=0 or looking at the graph

6-2 Solving Quadratic Equations by Graphing (cont.)

Graph with definitions shown: Three outcomes for number of roots:

One root:Two roots

No roots:

Vertex (2., -5.)

Root Root

Axis of Symmetry

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

1

2

3

4

5

6

7

8

9

10

6-2 Example

-x2: quadratic term-2x: linear term8: constant term

Vertex:x=(-b/2a)x= -(-2/2(-1))x= 2/(-2)x= -1

Solve for y:

y= -x2 -2x + 8

y= -(-1)2 -(2)(-1) + 8

y= -(1) + 2 + 8

y= 9 Vertex is (-1, 9)

For y= -x2 -2x + 8 identify each term, graph the equation, find the vertex, and find the solutions of the equation.

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

Root (-4., 0.) Root (2., 0.)

Vertex (-1., 9.)

6-2 Example (cont.)

Find the roots for the Problem:

-x2 -2x + 8 = 0

(-x + 2)(x + 4) = 0

-x + 2 = 0 x + 4 = 0

-x = -2 x = -4

x = 2

(2, 0) and (-4, 0) are the roots.

6-2 Problems

1. Name the quadratic term, the linear term, and the constant term of y= -x2 + 4x.

2. Graph y= 4x2 – 2x + 1 and find its vertex and axis of symmetry.

3. Find the roots of y= x2 – 8x + 12.

1)–x2: quadratic term 4x: linear term no constant term 2) (¼, ¾) x= ¼ 3) (2,0) and (6,0)

6-3 Solving Quadratic Equations by Factoring

• Factor with the zero product property: if a*b=0 then either a=0 or b=0 or both are equal to 0

• Factoring by guess and check is useful, but you may have to try several combinations before you find the correct one

• While doing word problems examine your solutions carefully to make sure it is a reasonable answer

6-3 Example

Solve the equation (2t + 1)2 – 4(2t + 1) + 3 = 0.

(2t + 1)(2t + 1) – 4(2t + 1) + 3 = 04t2 + 2t + 2t + 1 – 8t – 4 + 3 = 0

4t2 – 4t = 04t (4t – 1) = 0

4t = 0 t – 1 = 0t = 0 t = 1

The solutions are 0 and 1.

6-3 Problems

1. Solve (5x – 25)(7x + 3) = 0.2. Solve by factoring: 4x2 – 13x = 12.

1) 5 and -3/7 2) -3/4 and 4

6-4 Completing the Square

• The way to complete a square for x2 + bx + ? is to take ½ x b and then square it

• So for x2 + 6x + ? :½ (6) = 3 32 = 9 Therefore, the blank should be 9.

• If the coefficient of x2 is not 1, you must divide the equation by that coefficient before completing the square

• Some roots will be irrational or imaginary numbers

6-4 Example

Find the exact solution of 2x2 – 6x – 5 = 0.

2x2 – 6x – 5 = 0x2 – 3x – 5/2 = 0x2 – 3x + o = 5/2 + ox2 – 3x + 9/4 = 5/2 + 9/4(x – 3/2)2 = 19/4(x – 3/2)2 = 19/4

x – 3/2 = + 19/2 or

x – 3/2 = - 19/2

Solution:

x = 3/2 + 19/2 and

x = 3/2 – 19/2

6-4 Problems

1. Find the value c that makes x2 + 12x + c a perfect square.

2. Solve x2 – 2x – 15 = 0 by completing the square.

1) c = 36 2) -3 and 5

6-5 The Quadratic Formula and the Discriminant

• The quadratic formula gives the solutions of ax2 + bx + c = 0 when it is not easy to factor the quadratic or complete the square

• Quadratic formula:

• To remember the formula try singing it to the tune of the Notre Dame fight song or “Pop Goes the Weasel”

x = -b +/- b2 – 4ac

2a

6-5 The Quadratic Formula and the Discriminant (cont.)

• The b2 – 4ac term is called the discriminant and it helps to determine how many and what kind of roots you see in the solution

Value of b2 – 4ac Is it a perfect square? Nature of the Roots

b2 – 4ac > 0 yes 2 real roots, rational

b2 – 4ac > 0 no 2 real roots, irrational

b2 – 4ac < 0 not possible 2 imaginary roots

b2 – 4ac = 0 not possible 1 real root

6-5 Example

Find the discriminant of 3x2 + x – 2 = 0 and tell the nature of its roots. Then solve the equation.

Discriminant = b2 – 4ac = 12 – 4(3)(-2) = 1 – (-24) = 1 + 24

= 25

So, there are two real roots and the solutions will be rational.

a = 3 b = 1 c = -2

x = -1 +/- 12 – 4(3)(-2)

2(3)

x = -1 +/- 5 6

x = -1 + 5 x = -1 - 5 6 6

x = 2/3 x = -1

The solutions are 2/3 and -1.

x = -1 +/- 25 6

6-5 Problems

1. Use the discriminant to tell the nature of the roots of -7x2 – 8x – 10 = 0.

2. Use the quadratic formula to solve the equation -15x2 – 8x – 1 = 0.

1) Discriminant = -216 2 imaginary roots 2) -1/3 and -1/5

6-5 Example

Write a quadratic equation from the given roots -4 and -2/3.

-4 + -2/3 = -14/3-4 x -2/3 = 8/3

a=3 b=14 c=83x2 + 14x + 8 = 0

6-5 Problems

1. Given the roots -1/3 and -1/5, write the quadratic equation.

2. Solve the equation x2 + 3x – 18 = 0 and check your answers using the sum and product of the roots.

1) 15x2 + 8x + 1 = 0 2) -6 and 3

6-6 vertex form

• A parabola has the equation y = a (x – h)2 + k• The coefficients a, h, and k can be changed to create

similar parabolas• Changing “k” moves the parabola up (k > 0) or down

(k < 0)• A change in “h” moves the parabola to the right (h >

0) or left (h < 0)• Changing “a” makes a parabola open upwards (a > 0)

or downwards (a < 0), and also tells if the parabola is wider ( IaI < 1) or narrower ( IaI > 1)

6-6 Example

k = 1 the graph moves up oneh = -3 the graph moves three to

the lefta = 2 the graph is narrower and

opens upward

Predict the shape of the parabola y = 2 (x+3)2 + 1 and graph it on a graphing calculator to check your answer.

6-6 Problem

1. Predict the shape of y = (x + 2)2 + 1 and graph the equation on a graphing calculator.

1) Moved up one and two to the left

6-6 Analyzing Graphs of Quadratic Functions

• For more information on figuring out the shape of graphs see the notes on 6-6 p.323

• The equation y = a (x – h)2 + k gives the vertex (h, k) and the axis of symmetry is x = h

• You can write the equation of a parabola if you know its vertex or if you know three points the parabola passes through

6-6 Examples

1. Write y = x2 + 6x – 3 in standard form and then name the vertex, axis of symmetry ,and direction of opening.

y = x2 + 6x – 3y + 3 + o = (x2 + 6x + o)y + 3 + 9 = (x2 + 6x + 9)y + 12 = (x + 3)2

y = (x + 3)2 – 12

Vertex: (-3, -12)

Axis of Symmetry: x = -3

The graph should open upwards.

6-6 Problems

1. Write y = x2 – 6x + 11 in the form y = a (x – h)2 + k and find the vertex, axis of symmetry, and direction of opening.

1)y = (x – 3)2 + 2 vertex: (3, 2) axis of symmetry: x = 3 opens upward

2)y = 2x2 - x

6-7 Graphing and Solving Quadratic Inequalities

• The graph of the parabola serves as a boundary between the area inside the parabola and the area outside the parabola

• Graph quadratic inequalities the same way you graph linear inequalities:

• Graph the parabola and decide if the boundary line should be solid (≤ or ≥) or dashed ( < or >)

• Test one point inside the parabola and one outside the parabola• Shade the region where the inequality was true for the tested

points

• To solve a quadratic inequality you could graph it or find it through factoring the inequality and testing points

6-7 Examples

1. Graph the quadratic inequality y > 3x2 + 12x. Then decide if (2,4) is a solution to the inequality.

Decide where to shade:

Test: (0,0) Test: (-2, 2)

0 > 3 (0)2 + 12 (0) 2 > 3 (-2)2 + 12 (-2)

0 > 0 + 0 2 > 3 (4) – 24

0 > 0 2 > -12

False True

Is (2, 4) a solution?

4 > 3 (2)2 + 12(2) You could also look at the graph and see that

4 > 12 + 24 (2,4) is not in the shaded region.

4 > 36 (2, 4) is not a solution.

6-7 Problems

1. Graph the quadratic inequality y > x2 – x + 10 and decide if (0, 12) is a solution of the inequality.

2. Solve x2 – 10x – 16 < 0.

1)(0, 12) is a solution.2) 2 < x < 8