9.4 Formulas and Further Applications
To solve a formula for a squared variable, proceed as
follows.
(a) If the variable appears only to the second power:
Isolate the squared variable on one side of the equation,
and then use the square root property.
(b) If the variable appears to the first and second
powers:Write the equation in standard form, and then
use the quadratic formula.
Solve for r.
Multiply by
Divide by A.
Square root
property
Rationalize
r= denominator.
22 mpA
A
r=
B
2 mp
A
r^2 =
2 mp
A
r^2 A= 2 mp r^2.
A=
2 mp
r^2
CONCEPTS EXAMPLES
Solve for x.
Standard
form
x=
- r 2 r^2 + 4 t
2
a=1,b=r,c=-t
x=
- r 2 r^2 - 41121 - t 2
2112
x^2 +rx-t= 0
x^2 +rx=t
9.5 Graphs of Quadratic Functions
1.The graph of the quadratic function defined by
, is a parabola with
vertex at and the vertical line as axis.
2.The graph opens up if ais positive and down if ais
negative.
3.The graph is wider than the graph of if
06 |a| 61 and narrower if |a| 7 1.
ƒ 1 x 2 =x^2
1 h, k 2 x=h
F 1 x 2 a 1 xh 22 k,aZ 0
Graph
The graph opens down since
Vertex:
Axis:
Domain:
Range: 1 - q, 1 4
1 - q, q 2
x=- 3
1 - 3, 1 2
a 6 0.
x
y
0
1
–2
–5 1
(–3, 1)
x = –3
ƒ 1 x 2 =- 1 x+ 322 +1.
9.6 More about Parabolas and Their
Applications
The vertex of the graph of
may be found by completing the square.
The vertex has coordinates
Graphing a Quadratic Function
Step 1 Determine whether the graph opens up or
down.
Step 2 Find the vertex.
Step 3 Find the x-intercepts (if any). Find the y-
intercept.
Step 4 Find and plot additional points as needed.
Horizontal Parabolas
The graph of
or
is a horizontal parabola with vertex and the
horizontal line as axis. The graph opens to the right
if and to the left if
Horizontal parabolas do not represent functions.
a 70 a 6 0.
y=k
1 h, k 2
xay^2 byc xa 1 yk 22 h
A 2 ab , ƒA 2 abBB.
aZ0,
ƒ 1 x 2 ax^2 bxc, Graph
The graph opens up since
Vertex:
The solutions of are
and so the x-intercepts are
and
so the y-intercept is
Domain:
Range:
Graph
The graph opens to the right
since
Vertex:
Axis:
Domain:
Range: 1 - q, q 2
C^12 , qB
y=-^32
A^12 , -^32 B
a 7 0.
x
y
0
(^11)
–1^5
1
2
3
(), – 2
3
y = – 2
x= 2 y^2 + 6 y+5.
3 - 1, q 2
1 - q, q 2
ƒ 102 =3, 1 0, 3 2.
1 - 3, 0 2.
- 3, 1 - 1, 0 2
x^2 + 4 x+ 3 = 0 - 1
1 - 2, - 12
a 7 0.
x
y
0
3
–3 1
(–2, –1)
x = –2
–1
ƒ 1 x 2 =x^2 + 4 x+3.
CHAPTER 9 Summary 561
(continued)