Transcript 3.1 Powerpoint

Chapter 3.1
Quadratic Functions and Models
Example 1 Using Operations on Functions
Let f(x) = x2 + 1 and g(x) = 3x + 5
f
(d )  0
g
f ( x)  4x  x  2
f ( x)  5 x  1
2
f ( x)  x  2 x  4x
4
3
2
where a 3 is 2,
a 2 is - 12 ,
a 0 is 5.
T hepolynomial functions defined by
f(x)  x  2 x  4 x and
4
3
2
f(x)  4x  x  2 havedegrees 4 and 2
respectively.
2
The number an is the leading coefficient of f(x).
The function defined by f(x) – 0 is called the zero
polynomial. The zero polynomial has no degree.
However, a polynomial function defined by f(x) =
a0, for a nonzero number a0 has degree 0.
Quadratic Functions
In sections 2.3 and 2.4, we discussed first-degree
(linear) polynomial functions, in which the
greatest power of the variable is 1.
Now we look at polynomial functions of degree 2,
called quadratic functions.
The simplest quadratic function id given by
f(x) = x2 with a = 1, b = 0, and c = 0.
To find some points
on the graph of this
function, choose
some values for x and
find the
corresponding values
for f(x), as in the
table with Figure 1.
Then plot these
points, and draw a
smooth curve
through them.
This graph is called
a parabola.
Every quadratic
function has a graph
that is a parabola.
We can determine
the domain and the
range of a
quadratic function
whose graph is a
parabola, such as
the one in Figure 1,
from its graph.
Since the graph
extends indefinitely
to the right and to
the left, the domain
is (-∞, +∞).
Since the lowest
point is (0,0), the
minimum range
value (y-value) is 0.
The graph extends
upward indefinitely;
there is no
maximum y-value,
so the range is
[0, ∞).
Parabolas are
symmetric with
respect to the line
(the y-axis in Figure
1).
The line of
symmetry for a
parabola is called
the axis of the
parabola.
The point where the axis
intersects the parabola is called
the vertex of the parabola.
As Figure 2 shows,
the vertex of a parabola that
opens down is the highest
point of the graph
and the vertex of a parabola
that opens up is the lowest
point of the graph.
Graphing Techniques
The graphing techniques of Sections 2.6 applied to
the graph of f(x) = x2 give the graph of any
quadratic function.
The graph of g(x) = ax2
is a parabola with vertex
at the origin
that opens up if a is
positive
and downward if a is
negative.
The width of the graph of g(x) is determined by
the magnitude of a. That is, the graph of g(x) is
narrower than that of f(x) = x2 if |a| > 1
The width of the graph of g(x) is determined by
the magnitude of a. That is, the graph of g(x) is
narrower than that of f(x) = x2 if |a| > 1 and is
broader than that of f(x) = x2 if |a| < 1. By
completing the square, any quadratic can be
written in the form
F(x) = a(x-h)2 + k.
The width of the graph of g(x) is determined by
the magnitude of a. That is, the graph of g(x) is
narrower than that of f(x) = x2 if |a| > 1 and is
broader than that of f(x) = x2 if |a| < 1. By
completing the square, any quadratic can be
written in the form
F(x) = a(x-h)2 + k.
The graph of F(x) is the same as the graph of
g(x) = ax2 translated |h| units horizontally (to the
right is h is positive and to the left is h is negative)
and translated |k| units vertically
(up if k is positive and down if is k is negative).
Example 1 Graph each function. Give the
domain and the range
y
Graph each
function
f x   x 2  4 x  2
x
-1
x0
1
2
3
4
5
f(x)
Example 1 Graph each function. Give the
domain and the range
y
Graph each
function
f x   x 2  4 x  2
x
-1
x0
1
2
3
4
5
f(x)
Think of g(x) = -½x2
as g(x) = -(½x2) . The
graph of y = ½x2 is a
broader version of
y = x2,
Think of g(x) = -½x2
as g(x) = -(½x2) . The
graph of y = ½x2 is a
broader version of
y = x2,
and the graph of
g(x) = -(½x2) is a
reflection of y = ½x2
across the x-axis.
The vertex is (0,0), and
the axis of the parabola
is the line
x = 0 (the y-axis).
The domain is (- ∞, ∞),
and the range is (- ∞, 0].
We can write F(x) = -½(x-4)2 +3
as F(x) = g(x-h)2 + k, where
g(x) is the function of part (b),
h is 4, and k is 3.
Therefore the graph of F(x) is the
graph of g(x) granslated 4 usints
to the right and 3 units up.
See figure 7.
Therefore the graph of F(x) is the
graph of g(x) granslated 4 usints
to the right and 3 units up.
See figure 7.
The vertex is (4,3), and
the axis of the parabola
is the line
x = 4.
The domain is (- ∞, ∞),
and the range is (- ∞, 3].
Completing the Square
In general, the graph of the quadratic function
defined by
f(x) = a (x – h)2 + k
is a parabola with vertex (h, k) and axis x = h.
The parabola opens up if a is positive and
downward if a is negative.
With these facts in mind, we complete the square t
graph a quadratic function defined by
f(x) = ax2 + bx + c
Example 2 Getting a Parabola by Completing the Square.
Graph f(x) x - 6x  7 by completing
2
thesquare and locatingthe vertex.
Example 2 Graphing a Parabola by Completing the Square.
Graph f(x) x - 6x  7 by completing
2
thesquare and locatingthe vertex.
Example 3 Getting a Parabola by Completing the Square.
Graph f(x) - 3x - 2x  1 by completing
2
thesquare and locatingthe vertex.
Example 3 Getting a Parabola by Completing the Square.
Graph f(x) - 3x - 2x  1 by completing
2
thesquare and locatingthe vertex.
The Vertex Formula
We can generalize the earlier work to obtain a
formula for the vertex of a parabola. Starting with
the general quadratic form
f(x) = ax2 + bx + c
and completing the square will change the form to
f(x) = a (x – h)2 + k
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
f(x) = ax2 + bx + c
Comparing the last result with
f(x) = ax2 + bx + c
shows that
2
b
b
x
and k  c 
2a
aa
Example 4 Finding the Axis and the Vertex of a Parabola Using the
Formula
Find theaxis and vertexof theparabola having
equation
f(x) 2x  4x  5 using theformula.
2
Quadratic Models and Curve Fitting
From the graphs in this section, we see that
quadratic functions make good models for data
sets where the data either
increases, levels off, and then decreases
or
decreases, levels off, and then increases
Since the vertex of a vertical parabola is the
highest or lowest point on the graph, equations of
the form
y = ax2 + bx + c are important in problems where
we must find the maximum or minimum value of
some quantity. When a < 0, the y-coordinate of the
vertex gives the maximum value of y and the xvalue tells where it occurs. Similarly, when a>0,
the y-coordinate of the vertex gives the minimum
y-value.
An application of quadratic functions models the
height of a propelled object as a function of the
time elapsed after it is propelled. Recall that if air
resistance is neglected, the height s (in feet) of an
object propelled directly upward from an initial
height s0 feet with initial velocity v0 feet per
second is
s(t) = -16t2 +v0t + x0
where t is the number of seconds after the object is
propelled. The coefficient of t2 (that is -16) is a
constant based on the gravitational force of Earth.
This constant varies on other surfaces, such as the
moon or the other planets.
Example 5 Solving a Problem Involving Projectile Motion
A ball is thrown directly upward from an intial
height of 100 ft with an initial velocity of 80 ft
per sec.
(a) Give the function that describes the height of
the ball in terms of time t.
(b) Graph this function on a graphing calculator so
that the y-intercept, the positive x-intercept,
and the vertex are visible.
Example 5 Solving a Problem Involving Projectile Motion
(c) Figure 11 shows that the point (4.8, 115.36) lies
on the graph of the functions. What does this
mean for this particular situation.
(d) After how many seconds does the projectile
reach its maximum height? What is this
maximum height?
Example 5 Solving a Problem Involving Projectile Motion
(e) For what interval of time is the height of the
ball greater than 160 ft?.
(f) After how many seconds will the ball hit the
ground?
In Section 2.4 we introduced curve fitting and used
linear regression to determine linear equations that
modeled data. With a graphing calculator, we can
use a technique called quadratic regression to find
quadratic equations that model data.
Example 6. Modeling the Number of Hospital Outpatient Visits
The number of hospital outpatient visits (in
millions) for selected years is shown in the table.
In the table, 75 represents 1975,
85 represents 1985,
100 represents 2000, and so on,
the number of outpatient visits
is given in millions.
Example 6. Modeling the Number of Hospital Outpatient Visits
(a) Prepare a scatter diagram, and determine a
quadratic model for these data.
(b) Use the model from part (a) to predict the
number of visits in 2005.