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.