Transcript Chapter 6 Exploring Quadratic Functions and Inequalities

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= -10
Xmax=10
Ymin= -10
Ymax= 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.)
Graph with definitions shown: Three outcomes for number of
roots:
Axis of Symmetry
Two roots
Root
One root:
10
10
8
8
6
6
4
4
2
Root
-10
-8
-6
-4
-2
2
2
4
6
8
10
-10
-8
-6
-4
-2
2
-2
-2
-4
-4
-6
-6
-8
-8
-10
-10
Vertex (2., -5.)
No roots:
10
9
8
7
6
5
4
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
1
2
3
4
5
6
7
8
9
10
4
6
8
10
6-1 Example
For y= -x2 -2x + 8 identify each term, graph the equation, find the vertex, and
find the solutions of the equation.
10
-x2:
quadratic term
-2x: linear term
8: constant term
Vertex:
x=(-b/2a)
x= -(-2/2(-1))
x= 2/(-2)
x= -1
Vertex (-1., 9. )
8
6
4
2
Root (-4. , 0.)
Solve for y:
-10
-8
-6
-4
-2
2
-2
-4
y= -x2 -2x + 8
-6
-8
y=
-(-1)2
-(2)(-1) + 8
-10
y= -(1) + 2 + 8
y= 9
Root (2., 0. )
Vertex is (-1, 9)
4
6
8
10
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 = 0
4t2 + 2t + 2t + 1 – 8t – 4 + 3 = 0
4t2 – 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: 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 = 0
x2 – 3x – 5/2 = 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
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: x = -b +/- b2 – 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 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.
a = 3 b = 1 c = -2
Discriminant = b2 – 4ac
= 12 – 4(3)(-2)
= 1 – (-24)
= 1 + 24
= 25
So, there are two real roots
and the solutions will be
rational.
-1 +/- 12 – 4(3)(-2)
2(3)
x=
x=
x=
x = -1 + 5
6
x = 2/3
-1 +/- 25
6
-1 +/- 5
6
x = -1 - 5
6
x = -1
The solutions are 2/3 and -1.
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=8
3x2 + 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
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.
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 – 3
y + 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 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:
Plug in c = 1 for the other
two equations:
(0, 1): 1 = a(0)2 + b (0) + c
-1 = 4a + 2b + 1
1=c
-2 = 4a + 2b
(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
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.
2) y = 2x2 - x
1) y = (x – 3)2 + 2
opens upward
Graph
of #2
vertex: (3, 2) axis of symmetry: x = 3
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)
0 > 3 (0)2 + 12 (0)
0>0+0
Test: (-2, 2)
2 > 3 (-2)2 + 12 (-2)
2 > 3 (4) – 24
0>0
2 > -12
False
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 x2 – 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
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.
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}.
3 + 5 + 6 + 7 + 9 + 11 + 22 = 9
Mean =
7
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
3σ
2σ
1σ
1σ
2σ 3σ
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
2.5%
50 53
95%
56 59
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.