1. Chapter 6 Exploring Quadratic Functions and Inequalities

2. 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

3. 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

4. 6-2 Example Graph y= -x2 - 2x + 8 and find its roots. Vertex: (-1, 9)
Viewing window:
Xmin= -10
Xmax=10
Ymin= -10
Ymax= 10

5. 6-2 Problems
Find what size viewing window is needed to view y= x2 + 4x Find the roots.
Window: Xmin= -10 Xmax= 10 Ymin= -20 Ymax= Roots: and

6. 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

7. 6-2 Solving Quadratic Equations by Graphing (cont.)
Graph with definitions shown:
Three outcomes for number of roots:
Two roots
One root:
No roots:

8. 6-2 Example
For y= -x2 -2x + 8 identify each term, graph the equation, find the vertex, and find the solutions of the equation.
-x2: quadratic term
-2x: linear term
8: 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)
y= 9
Vertex is (-1, 9)

9. 6-2 Example (cont.) Find the roots for the Problem: -x2 -2x + 8 = 0
(2, 0) and (-4, 0) are the roots.

10. 6-2 Problems
Name the quadratic term, the linear term, and the constant term of y= -x2 + 4x.
Graph y= 4x2 – 2x + 1 and find its vertex and axis of symmetry.
Find the roots of y= x2 – 8x + 12.
–x2: quadratic term 4x: linear term no constant term 2) (¼, ¾) x= ¼ ) (2,0) and (6,0)

11. 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

12. 6-3 Example Solve the equation (2t + 1)2 – 4(2t + 1) + 3 = 0.
(2t + 1)(2t + 1) – 4(2t + 1) + 3 = 0
4t2 + 2t + 2t + 1 – 8t – = 0
4t2 – 4t = 0
4t (4t – 1) = 0
4t = t – 1 = 0
t = t = 1
The solutions are 0 and 1.

13. 6-3 Problems Solve (5x – 25)(7x + 3) = 0.
Solve by factoring: 4x2 – 13x = 12.
1) 5 and -3/ ) -3/4 and 4

14. 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) = = 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

15. 6-4 Example Find the exact solution of 2x2 – 6x – 5 = 0.
x2 – 3x + o = 5/2 + o
x2 – 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

16. 6-4 Problems
Find the value c that makes x2 + 12x + c a perfect square.
Solve x2 – 2x – 15 = 0 by completing the square.
1) c = ) -3 and 5

17. 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

18. 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 no
2 real roots, irrational
b2 – 4ac < 0
not possible
2 imaginary roots
b2 – 4ac = 0
1 real root

19. 6-5 Example
Find the discriminant of 3x2 + x – 2 = 0 and tell the nature of its roots. Then solve the equation.
a = 3 b = 1 c = -2
x = / – 4(3)(-2)
Discriminant = b2 – 4ac = 12 – 4(3)(-2) = 1 – (-24) = = 25
So, there are two real roots and the solutions will be rational.
2(3)
x = /
x = /
x = x =
x = 2/ x = -1
The solutions are 2/3 and -1.

20. 1) Discriminant = -216 2 imaginary roots 2) -1/3 and -1/5
6-5 Problems
Use the discriminant to tell the nature of the roots of -7x2 – 8x – 10 = 0.
Use the quadratic formula to solve the equation -15x2 – 8x – 1 = 0.
1) Discriminant = imaginary roots ) -1/3 and -1/5

21. 6-5 Example
Write a quadratic equation from the given roots -4 and -2/3.
/3 = -14/3
-4 x -2/3 = 8/3
a=3 b=14 c=8
3x2 + 14x + 8 = 0

22. 6-5 Problems
Given the roots -1/3 and -1/5, write the quadratic equation.
Solve the equation x2 + 3x – 18 = 0 and check your answers using the sum and product of the roots.
1) 15x2 + 8x + 1 = ) -6 and 3

23. 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)

24. 6-6 Example
Predict the shape of the parabola y = 2 (x+3)2 + 1 and graph it on a graphing calculator to check your answer.
k = 1 the graph moves up one
h = -3 the graph moves three to the left
a = 2 the graph is narrower and opens upward

25. 1) Moved up one and two to the left
6-6 Problem
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

26. 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

27. 6-6 Examples
Write y = x2 + 6x – 3 in standard form and then name the vertex, axis of symmetry ,and direction of opening.
y = x2 + 6x – 3
y o = (x2 + 6x + o)
y = (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.

28. 6-6 Problems
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.
y = 2x2 - x
y = (x – 3) vertex: (3, 2) axis of symmetry: x = 3 opens upward

29. 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

30. 6-7 Examples
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) (0) > 3 (-2) (-2)
0 > > 3 (4) – 24
0 > > -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 > (2,4) is not in the shaded region.
4 > (2, 4) is not a solution.

31. 6-7 Problems
Graph the quadratic inequality y > x2 – x + 10 and decide if (0, 12) is a solution of the inequality.
Solve x2 – 10x – 16 < 0.
(0, 12) is a solution. 1)
2) 2 < x < 8