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Start- Up Day 10
Copyright © 2015, 2011, and 2007 Pearson Education, Inc.
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What you’ll learn about
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Polynomial Functions
Linear Functions and Their Graphs
Average Rate of Change
Association, Correlation, and Linear Modeling
Quadratic Functions and Their Graphs
Applications of Quadratic Functions
… and why
Many business and economic problems are modeled by
linear functions. Quadratic and higher-degree
polynomial functions are used in science and
manufacturing applications.
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2.1
Objective: Student will be able to recognize and
graph linear and quadratic functions, and use these
functions to model situations and solve problems.
Essential Question: What is a “rate of change”? What is the
relationship between a quadratic function and its graph? How
can we create linear/quadratic models to help us solve
problems?
Home Learning: pg. 169 4, 9, 13‐18, 21, 25 & 43
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Polynomial Function
Let n be a nonnegative integer and let a0 ,a1 ,a2 ,...,an1 ,an
be real numbers with an  0. The function given by
f (x)  an x n  an1 x n1  ...  a2 x 2  a1 x  a0
is a polynomial function of degree n.
The leading coefficient is an .

The zero function f x  0 is a polynomial function.
It has no degree and no leading coefficient.
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No and Low Polynomial Functions
Name
Zero Function
Form
f(x) = 0
Constant Function
f(x) = a (a ≠ 0)
0
Linear Function
f(x) = ax + b (a ≠ 0)
1
Quadratic Function f(x) = ax2 + bx + c (a ≠ 0)
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Degree
Undefined
2
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Example 1: Which are Polynomial Functions?
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Your Turn: Try #1-#6
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Example: Determining an Equation of a Linear
Function
Write an equation for the linear function f
such that f (1)  2 and f (2)  3.
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Solution:
The line contains the points (1,2) and (2, 3). Find the slope:
3 2 1
m

2 1 3
Use the point-slope formula and the point (2, 3):
y  y1  m(x  x1 )
1
y  3  x  2 
3
1
2
y3 x
3
3
1
7
y x
3
3
1
7
f (x)  x 
3
3
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Slope = Average Rate of Change
The average rate of change of a function
y  f (x) between x  a and x  b, a  b, is
f (b)  f (a)
.
ba
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Constant Rate of Change Theorem
A function defined on all real numbers is a linear
function if and only if it has a constant nonzero
average rate of change between any two points
on its graph.
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Your Turn: (#8) Write a linear
equation that satisfies the given
conditions & include a graph.
f (-3) = 5
f (6) = -2
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Linear Correlation?
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Properties of the
Correlation Coefficient, r
1. –1 ≤ r ≤ 1
2. When r > 0, there is a positive linear
association.
3. When r < 0, there is a negative linear
association.
4. When |r| ≈ 1, there is a strong linear
association..
5. When r ≈ 0, there is weak or no linear
association.
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Regression Analysis
1. Enter and plot the data (scatter plot).
2. Find the regression model that fits the
problem situation.
3. Superimpose the graph of the regression
model on the scatter plot, and observe the fit.
4. Use the regression model to make the
predictions called for in the problem.
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Vertex Form of a Quadratic Equation
Any quadratic function f(x) = ax2 + bx + c,
a ≠ 0, can be written in the vertex form
f(x) = a(x – h)2 + k
The graph of f is a parabola with vertex (h, k)
and axis x = h, where h = –b/(2a). If a > 0, the
parabola opens upward, and if a < 0, it opens
downward.
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The Graph of f(x)=ax2
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Example: Understanding
Transformations of a Quadratic
Function
Describe how to transform the graph of f (x)  x 2 into the
graph of f (x)  2 x  2   3.
2
The graph of f (x)  2 x  2   3
2
is obtained by vertically stretching
the graph of f (x)  x 2 by a factor
of 2 and translating the resulting
graph 2 units right and 3 units up.
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Example: Finding the Vertex and Axis
of a Quadratic Function
Use the vertex form of a quadratic function to find the
vertex and axis of the graph of f (x)  2x 2  8x  11.
Rewrite the equation in vertex form.
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Solution
The standard polynomial form of f is f (x)  2x 2  8x  11;
a  2, b  8, c  11, and the coordinates of the vertex are
b
8
h
  2 and k  f (h)  f (2)  2(2)2  8(2)  11  3.
2a 4
The equation of the axis is x  2, the vertex is (2,3), and the
vertex form of f is f (x)  2(x  2)2  3.
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Example: Completing the Square to describe
the graph of a Quadratic Function
Use completing the square to describe the graph of
f (x)  4x 2  12x  8.
Support your answer graphically.
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Solution
f (x)  4x 2  12x  8

 4 x

 4 x 2  3x  8
2
 3x     
 8
2
2
 2
 3  3 
 4  x  3x         8
 2  2 

 2
 4  x  3x 

9
 9
  4    8
4
4
2
3

 4  x    1

2
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Can You Write the Equation?
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Characterizing the Nature of
a Quadratic Function
Point of
View
Verbal
Characterization
Algebraic
f(x) = ax2 + bx + c or
f(x) = a(x – h)2 + k (a ≠ 0)
Graphical
parabola with vertex (h, k) and axis x = h;
opens upward if a > 0, opens downward if
a < 0; initial value = y-intercept = f(0) = c;
polynomial of degree 2
b  b 2  4ac
x-intercepts 
2a
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Your Turn:
(#30) Rewrite the equation in vertex form.
Identify the vertex and axis of the graph.
f (x) = 6 - 2x + 4x
2
(#34) Complete the Square to rewrite in vertex
form. Describe the graph.
2
f (x) = x - 6x +12
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Vertical Free-Fall Motion
The height s and vertical velocity v of an object in
free fall are given by
1 2
s(t) = - gt + v0 t + s0 and v(t) = -gt + v0 ,
2
2
2
where t is time (in seconds), g » 32 ft/sec » 9.8 m/sec
is the acceleration due to gravity, v0 is the initial
vertical velocity of the object, and s0 is its initial height.
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Let’s Try It: Vertical Free-Fall Motion
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