Chapter 3.1
Quadratic Functions and Models

Example 1 Using Operations on Functions
Let f(x) = x 2 + 1 and g(x) = 3x + 5
( d )

 f g


 

f ( x )

5 x

1 f ( x )
 x
4 
2 x
3 
4 x
2 f ( x )

4 x
2  x

2

where a
3 is 2, a
2 is 1
2
, a
0 is 5.


4 
x
3 
x
2

2  x


The number a n 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) = a
0
, for a nonzero number a
0 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) = x 2 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) = x 2 give the graph of any quadratic function.

The graph of g(x) = ax 2 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) = x 2 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) = x 2 if |a| > 1 and is broader than that of f(x) = x 2 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) = x 2 if |a| > 1 and is broader than that of f(x) = x 2 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) = ax 2 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 f
Graph each function
 x
2 
4 x

2 x
-1
0
1
2
3
4
5 x f(x)

Example 1 Graph each function. Give the domain and the range y f
Graph each function
 x
2 
4 x

2 x
-1
0
1
2
3
4
5 x f(x)

Think of g(x) = -½x 2 as g(x) = -(½x 2 ) . The graph of y = ½x 2 is a broader version of y = x 2 ,

Think of g(x) = -½x 2 as g(x) = -(½x 2 ) . The graph of y = ½x 2 is a broader version of y = x 2 , and the graph of g(x) = -(½x 2 ) is a reflection of y = ½x 2 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) = ax 2 + bx + c

Example 2 Getting a Parabola by Completing the Square.
Graph f(x)

x
2
6x

7 by completing the square and locating the vertex.

Example 2 Graphing a Parabola by Completing the Square.
Graph f(x)

x
2
6x

7 by completing the square and locating the vertex.

Example 3 Getting a Parabola by Completing the Square.
Graph f(x)

3x
2
2x

1 by completing the square and locating the vertex.

Example 3 Getting a Parabola by Completing the Square.
Graph f(x)

3x
2
2x

1 by completing the square and locating the 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) = ax 2 + bx + c and completing the square will change the form to f(x) = a (x – h) 2 + k

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

f(x) = ax 2 + bx + c

Comparing the last result with f(x) = ax 2 + bx + c shows that x
  b
2 a and k b
2
 c
 aa

Example 4 Finding the Axis and the Vertex of a Parabola Using the
Formula
Find the axis and vertex of the parabola having equation f(x)

2x
2 
4x

5 using the formula.

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 = ax 2 + 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 s
0 second is feet with initial velocity v
0 feet per s(t) = -16t 2 +v
0 t + x
0 where t is the number of seconds after the object is propelled. The coefficient of t 2 (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.