Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

The graph of any quadratic function is a parabola with a vertical axis.


NOTE We use the variable yand function notation interchangeably. Although


we use the letter ƒ most often to name quadratic functions, other letters can be used.


We use the capital letter Fto distinguish between different parabolas graphed on the


same coordinate axes.


Parabolas have a special reflecting property that makes


them useful in the design of telescopes, radar equipment, so-


lar furnaces, and automobile headlights. (See the figure.)


OBJECTIVE 2 Graph parabolas with horizontal and vertical shifts.Parabolas


need not have their vertices at the origin, as does the graph of


Graphing a Parabola ( Vertical Shift)

Graph


The graph of has the same shape as that of but is


shifted, or translated, 2 units down, with vertex Every function value is 2


less than the corresponding function value of Plotting points on both


sides of the vertex gives the graph in FIGURE 6.


ƒ 1 x 2 = x^2.


1 0, - 22.


F 1 x 2 =x^2 - 2 ƒ 1 x 2 = x^2


F 1 x 2 =x^2 - 2.


EXAMPLE 1


ƒ 1 x 2 =x^2.


ƒ 1 x 2


532 CHAPTER 9 Quadratic Equations, Inequalities, and Functions


Quadratic Function

A function that can be written in the form


for real numbers a, b, and c, with aZ 0,is a quadratic function.


ƒ 1 x 2 ax^2 bxc


Headlight


  • 2

  • 1
    0
    1
    2


2


  • 1

  • 2

  • 1
    2


4
1
0
1
4

x


– 220


2
x

y

f f (x) x^22


f f (x) x^22


F F (x) x^22 – – 22


F F (x) xx^22 – – 22


FIGURE 6

This parabola is symmetric about its axis so the plotted points are “mirror


images” of each other. Since xcan be any real number, the domain is still. The


value of y 1 or F 1 x 22 is always greater than or equal to -2,so the range is 3 - 2, q 2.


1 - q, q 2


x= 0,


Vertex:
Axis:
Domain:
Range:
The graph of
is shown for comparison.

ƒ 1 x 2 =x^2

3 - 2, q 2

1 - q, q 2

x= 0

1 0, - 22

F 1 x 2 =x^2 - 2

NOW TRY

NOW TRY
EXERCISE 1
Graph. Give
the vertex, axis, domain,
and range.


ƒ 1 x 2 =x^2 - 3

NOW TRY ANSWER






vertex: axis: ;
domain: ; range: 1 - q, q 2 3 - 3, q 2

1 0, - 32 ; x= 0

x

y

–2 0 2

1

6

–3
f(x) = x^2 – 3
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