chapter 6 exploring quadratic functions and inequalities
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Chapter 6
Exploring Quadratic Functions and Inequalities
By Jennifer Huss
6-1A Graphing Technology: 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
6-1A 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= -x2 - 2x + 8 and find its roots.
Vertex: (-1, 9)Roots: (-4, 0) (2, 0)Viewing window:Xmin= -10Xmax=10Ymin= -10Ymax= 10
6-1A 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
6-1 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
6-1 Solving Quadratic Equations by Graphing (cont.)
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
Graph with definitions shown:
-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
Three outcomes for number of roots:
One root:Two roots
No roots:
6-1 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)
-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.)
For y= -x2 -2x + 8 identify each term, graph the equation, find the vertex, and find the solutions of the equation.
6-1 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.
6-1 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)
6-2 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 = 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.
6-2 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
6-3 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
6-3 Example
Find the exact solution of 2x2 – 6x – 5 = 0.
2x2 – 6x – 5 = 0x2 – 3x – 5/2 = 0x2 – 3x + = 5/2 + 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
6-3 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
6-4 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
6-4 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
6-4 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
6-4 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 Sum and Product of Roots
• You can find the quadratic equation from the roots of the equation
• If the roots are called S1 and S2, then S1 + S2 = -b/a and S1 x S2 = c/a
• This gives us the coefficients of ax2 + 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=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
6-6A 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
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
Predict the shape of the parabola y = 2 (x+3)2 + 1 and graph it on a graphing calculator to check your answer.
6-6A 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 6-6A
• 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 = x2 + 6x – 3 in standard form and then name the vertex, axis of symmetry ,and direction of opening.
y = x2 + 6x – 3y + 3 + = (x2 + 6x + )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 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 = ax2 + 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 a = 2 – b
-2 = 4 (2 – b) + 2b a = 2 – 5
-2 = 8 – 4b + 2b a = -3
-2 = 8 – 2b
-10 = -2b a = -3 b = 5 c = 1
b = 5
The equation is y = -3x2 + 5x + 1.
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.
2. Find the equation of the parabola that passes through (0, 0), (2, 6) and (-1, 3). Then graph the function.
1)y = (x – 3)2 + 2 vertex: (3, 2) axis of symmetry: x = 3 opens upward
2)y = 2x2 - x
Graph of #2
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 > 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 Examples (cont.)
2. Solve x2 – 16 < 0.
(x – 4)(x + 4) = 0x = 4 and x = -4
Test in each region so lets choose x = -5, x = 0, and x = 5.Test: x = -5 Test: x = 0 Test: x = 5(-5)2 – 16 < 0 (0)2 – 16 < 0 (5)2 – 16 < 0 25 – 16 < 0 0 – 16 < 0 25 – 16 < 0 9 < 0 -16 < 0 9 < 0 False True False
The solution is -4 < x < 4.
-4 4
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
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 = (x1 – x)2 + (x2 – x)2 + … + (xn – 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
7= 5.8
6-8 Problem
1. Calculate the mean and standard deviation of {3, 5, 2, 6, 5, 9}.
1) Mean = 5 Standard Deviation = 6 or 2.45
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
41 44 47 53 56 5950
95% 2.5%2.5%
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.
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.
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