To derive a formula for the vertex of the graph of the quadratic function defined
by complete the square.
ƒ 1 x 2 = acx- a
- b
2 a
bd
2+
4 ac -b^2
4 a
= aax+
b
2 a
b
2+
4 ac- b^2
4 a
= aax^2 +
b
a
x+
b^2
4 a^2
b -
b^2
4 a
+ c
= aax^2 +
b
a
x+
b^2
4 a^2
b +aa-
b^2
4 a^2
b +c
= aax^2 +
b
a
x+
b^2
4 a^2
-
b^2
4 a^2
b +c
C^12 AbaBD(^2) =
A 2 baB
(^2) = b^2
4 a^2
= aax^2 +
b
a
xb +c
ƒ 1 x 2 = ax^2 + bx+ c
ƒ 1 x 2 =ax^2 +bx+ c 1 with aZ 02 ,
542 CHAPTER 9 Quadratic Equations, Inequalities, and Functions
Standard formFactor afrom the
first two terms.Add and subtractDistributive propertyFactor. Rewrite terms with
a common denominator.The vertex can be ex-
pressed in terms of a, b, and c.1 h, k 2ƒ 1 x 2 =a 1 x-h 22 +k- ab
2
4 a^2 =-b^2
4 ab^2
4 a^2.hkThe expression for kcan be found by replacing xwith Using function notation,
if then the y= ƒ 1 x 2 , y-value of the vertex is ƒA- 2 abB.
- b
2 a.
⎧⎨⎩ ⎧⎪⎨⎪⎩Vertex FormulaThe graph of the quadratic function defined by
has vertex
and the axis of the parabola is the line
x
b
2 a
.
a
b
2 a
, ƒa
b
2 a
bb,
aZ 02
ƒ 1 x 2 =ax^2 +bx+ c 1 with
Using the Formula to Find the VertexUse the vertex formula to find the vertex of the graph of
The x-coordinate of the vertex of the parabola is given by
- b
2 a
=
- 1 - 12
2112
=
1
2
- b
2 a.
ƒ 1 x 2 =x^2 - x-6.
EXAMPLE 3
and
x-coordinate of vertexa=1, b=-1, c=-6.The y-coordinate is
y-coordinate of vertexThe vertex is A NOW TRY
12 , -
254 B.
ƒa
1
2
b = a
1
2
b
2-
1
2
- 6 =
1
4
-
1
2
- 6 =-
25
4
ƒA- 2 abB =ƒA^12 B.
NOW TRY
EXERCISE 3
Use the vertex formula to find
the vertex of the graph of
ƒ 1 x 2 = 3 x^2 - 2 x+8.NOW TRY ANSWER
- A^13 ,^233 B