Applied Mathematics for Business and Economics

Lecture Note Function

y

3

2

1

½ 1 2 3 4 5 6 x

Comment

The graph of yf==++(xaxbx)^2 c is a parabola as long as. All

parabolas have a U shape, and

a≠ 0

( )

yf= xaxbx=++^2 copens either up

(if ) or down (if ). The “Peak” or “Valley” of the parabola is called

its vertex, and in either case, the x coordinate of the vertex is

a> 0 a< 0

2

b

x

a

=−.

Note that to get a reasonable sketch of the parabolay=ax^2 ++bx c, you need

only determine.

1 The location of the vertex

2 Whether the parabola opens up (a> 0 ) or down (a< 0 )

3 Any intercepts.

Example 3

For the equation y=xx^2 −+ 64

a. Find the Vertex.
b. Find the minimum value for y.
c. Find the x-intercepts.
d. Sketch the graph.
Solution

a. We haveab==− =1, 6 , a n d c 4. The vertex occurs at

6

3

21

x

−

=−=

×

Substituting x = 3 givesy= 362 −×+=− 345. The vertex is(3, 5− ).

b. Since a=> 10 and the parabola opens upward, y=− 5 is the minimum

value for y.

c. The x-intercept are found by setting xx^2 − 64 += 0 and solving for x

63616

3

2

x

±−

==± 5

d. The graph opens upward becausea=^10 > .The vertex is(3, 5− )

The axis of symmetry isx= 3.

The x-intercepts arex=± 3 5.