By Jennifer Huss

6-1A Graphing Technology: Quadratic
Functions
• Functions with the form y=ax 2 +bx+c are called quadratic functions and their graphs have a parabolic shape
• When we solve ax 2 +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

6-1A Graphing Technology: 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

6-1A Example
Graph y= -x 2 - 2x + 8 and find its roots.
Vertex: (-1, 9)
Roots: (-4, 0) (2, 0)
Viewing window:
Xmin= -10
Xmax=10
Ymin= -10
Ymax= 10

6-1A Problems
1. Find what size viewing window is needed to view y= x 2 + 4x -15. Find the roots.
and 2.3589
: -6.3589 Roots max= 10 Y n= -20 Ymi = 10 Xmax -10 in= : Xm Window

6-1 Solving Quadratic Equations by
Graphing
• In a quadratic equation y=ax 2 +bx+c, ax 2 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

6-1 Solving Quadratic Equations by
Graphing (cont.)
Graph with definitions shown:
Three outcomes for number of roots:
Axis of Symmetry
Root Root
Vertex (2., -5.)
Two roots
10
8
6
4
2
-6 -4 2 4 -10 -8 -2
-2
-4
-6
-8
-10
6 8 10
One root:
10
4
2
8
6
-10 -8 -6 -4 -2
-2
-4
-6
-8
-10
2 4 6 8 10
No roots:
10
3
2
1
5
4
9
8
7
6
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

6-1 Example
For y= -x 2 -2x + 8 identify each term, graph the equation, find the vertex, and find the solutions of the equation.
-x 2 : 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= -x 2 -2x + 8 y= -(-1) 2 -(2)(-1) + 8 y= -(1) + 2 + 8 y= 9
-10
Root (-4., 0.)
-8 -6 -4 -2
-8
-10
-2
-4
-6
10
Vertex (-1., 9.)
8
6
4
2
Root (2., 0.)
2 4 6 8 10
Vertex is (-1, 9)

Find the roots for the Problem:
-x 2 -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.

6-1 Problems
1. Name the quadratic term, the linear term, and the constant term of y= -x 2 + 4x.
2. Graph y= 4x 2 – 2x + 1 and find its vertex and axis of symmetry.
3. Find the roots of y= x 2 – 8x + 12.
(2,0) and (6,0) 3) ¼ x= (¼, ¾) 2) term onstant no c : linear term 4x c term : quadrati
2
–x
1)

6-2 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

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

6-2 Problems
1. Solve (5x – 25)(7x + 3) = 0.
2. Solve by factoring: 4x 2 – 13x = 12.
and 4 2) -3/4 and -3/7 1) 5

6-3 Completing the Square
• The way to complete a square for x 2 + bx + ? is to take ½ x b and then square it
• So for x 2 + 6x + ? :
½ (6) = 3 3 2 = 9 Therefore, the blank should be 9.
• If the coefficient of x 2 is not 1, you must divide the equation by that coefficient before completing the square
• Some roots will be irrational or imaginary numbers

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

6-3 Problems
1. Find the value c that makes x 2 + 12x + c a perfect square.
2. Solve x 2 – 2x – 15 = 0 by completing the square.
and 5 -3 2) = 36 1) c

6-4 The Quadratic Formula and the
Discriminant
• The quadratic formula gives the solutions of ax 2 + bx + c = 0 when it is not easy to factor the quadratic or complete the square
• Quadratic formula: x = -b +/- b 2 – 4ac
2a
• To remember the formula try singing it to the tune of the Notre Dame fight song or “Pop Goes the Weasel”

6-4 The Quadratic Formula and the
Discriminant (cont.)
• The b 2 – 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 b 2 – 4ac b2 – 4ac > 0 b2 – 4ac > 0 b2
– 4ac < 0 b2 – 4ac = 0
Is it a perfect square?
yes no not possible not possible
Nature of the Roots
2 real roots, rational
2 real roots, irrational
2 imaginary roots
1 real root

6-4 Example
Find the discriminant of 3x 2 + x – 2 = 0 and tell the nature of its roots. Then solve the equation.
Discriminant = b 2 – 4ac
= 1 2 – 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 +/- 1 2 – 4(3)(-2)
2(3) x = -1 +/- 25
6 x = -1 +/- 5
6 x = -1 + 5 x = -1 - 5
6 6 x = 2/3 x = -1
The solutions are 2/3 and -1.

6-4 Problems
1. Use the discriminant to tell the nature of the roots of -7x 2 – 8x – 10 = 0.
2. Use the quadratic formula to solve the equation -15x 2 – 8x – 1 = 0.
-1/5 1/3 and 2) nary roots 2 imagi = -216 riminant sc 1) Di

6-5 Sum and Product of Roots
• You can find the quadratic equation from the roots of the equation
• If the roots are called S
1
S
1
+ S
2
= -b/a and S
1 and S x S
2
2
, then
= c/a
• This gives us the coefficients of ax 2 + bx + c = 0
• You can also use this method with imaginary roots or to check your solution to a quadratic equation

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=8
3x 2 + 14x + 8 = 0

6-5 Problems
1. Given the roots -1/3 and -1/5, write the quadratic equation.
2. Solve the equation x 2 + 3x – 18 = 0 and check your answers using the sum and product of the roots.
and 3 2) -6 1 = 0 + + 8x
2
1) 15x

6-6A Graphing Technology: Families of
Parabolas
• 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)

6-6A 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

6-6A Problem
1. Predict the shape of y = (x + 2) 2 + 1 and graph the equation on a graphing calculator.
the left o to ed up one and tw 1) Mov

6-6 Analyzing Graphs of Quadratic
Functions
• For more information on figuring out the shape of graphs see the notes on 6-6A
• 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 = x 2 + 6x – 3 in standard form and then name the vertex, axis of symmetry ,and direction of opening.
y = x 2 + 6x – 3 y + 3 + o = (x 2 + 6x + o ) y + 3 + 9 = (x 2 + 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 Examples (cont.)
2.
Given the points (0, 1) (2, -1) and (1, 3) write the equation of the parabola.
Substitute the points into the equation y = ax 2 + bx + c:
(0, 1): 1 = a(0) 2 + b (0) + c
1 = c
(2, -1): -1 = a (2) 2 + b (2) + c
-1 = 4a + 2b + c
(1, 3): 3 = a(1) 2 + b (1) + c
3 = a + b + c
Plug in c = 1 for the other two equations:
-1 = 4a + 2b + 1
-2 = 4a + 2b
3 = a + b + 1
2 = a + b

6-6 Examples (cont.)
2.
Now solve the system of equations:
-2 = 4a + 2b
2 = a + b a = 2 – b
-2 = 4 (2 – b) + 2b
-2 = 8
– 4b + 2b
-2 = 8 – 2b
-10 = -2b b = 5 a = 2 a = 2 a = -3
– b
– 5 a = -3 b = 5 c = 1
The equation is y = -3x 2 + 5x + 1.

6-6 Problems
1. Write y = x 2 – 6x + 11 in the form y = a (x – h) 2 + k and find the vertex, axis of symmetry, and direction of opening.
2. Find the equation of the parabola that passes through (0, 0), (2, 6) and (-1, 3). Then graph the function.
of #2
Graph x = 3 y: mmetr of sy is ax (3, 2) ertex: v ard
+ 2
2
-x
2
2x
(x –3) opens upw y = 2) y = 1)

6-7 Graphing and Solving Quadratic
Equations
• 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 > 3x 2 + 12x. Then decide if (2,4) is a solution to the inequality.
Decide where to shade:
Test: (0,0)
0 > 3 (0) 2 + 12 (0)
0 > 0 + 0
0 > 0
False
Test: (-2, 2)
2 > 3 (-2) 2 + 12 (-2)
2 > 3 (4) – 24
2 > -12
True
Is (2, 4) a solution?
4 > 3 (2) 2 + 12(2)
4 > 12 + 24
4 > 36
You could also look at the graph and see that
(2,4) is not in the shaded region.
(2, 4) is not a solution.

6-7 Examples (cont.)
2.
Solve x 2 – 16 < 0.
(x – 4)(x + 4) = 0 x = 4 and x = -4
-4 4
Test in each region so lets choose x = -5, x = 0, and x = 5.
Test: x = -5
(-5) 2 – 16 < 0
25 – 16 < 0
Test: x = 0
(0) 2 – 16 < 0
0 – 16 < 0
Test: x = 5
(5) 2 – 16 < 0
25 – 16 < 0
9 < 0
False
-16 < 0
True
9 < 0
False
The solution is -4 < x < 4.

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

6-8 Integration: Statistics- Standard
Deviation
• Standard deviation tells how spread out the values are in a set of data (given symbol σ)
• The mean is the average of your data ( symbol x )
• Usually a graphing calculator is used to calculate the standard deviation
Standard
Deviation
=
(x
1
– x) 2 + (x
2
– x) 2 + … + (x n
– x) 2 n

6-8 Example
Calculate the mean and standard deviation of
{3, 5, 6, 7, 9, 11, 22}.
Mean =
3 + 5 + 6 + 7 + 9 + 11 + 22
7
= 9
Standard
=
Deviation
(3 – 9) 2 + (5 – 9) 2 + … + (22 - 9) 2
= 5.8
7

6-8 Problem
1. Calculate the mean and standard deviation of
{3, 5, 2, 6, 5, 9}.
2.45
6 or on = Deviati Standard = 5 1) Mean

6-9 Integration: Statistics – Normal
Distribution
• A normal distribution curve shows the frequency
(how many times something occurs) in a symmetric graph
– It is often called a bell-curve because it resembles a bell
• Normal Distributions have the following properties:
1. The graph is the highest at the mean
2. The mean, median, and mode are equal
3. Data is symmetrical about the mean

6-9 Integration: Statistics – Normal
Distribution
For a Normal Distribution curve:
•68% of the values fall within one standard deviation
•95% of the values fall within two standard deviations
•99% of the values fall within three standard deviations
3 σ 2 σ 1 σ 1 σ 2 σ 3 σ

6-9 Example
• A battery has an average life span of 50 hours, with a standard deviation of 3 hours. The life span of the batteries is normally distributed.
a) What percent of batteries last at least 44 hours?
97.5% of batteries last at least 44 hours.
41 44 47 50 53 56 59
2.5% 95% 2.5% b) If we have 1500 batteries, how many batteries are within one standard deviation of the mean?
68% of batteries are within one standard deviation.
(1500)(0.68) = 1020 1020 batteries are within one standard deviation.