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Let’s check out your problem:
Find the
a
a
a
value of a parabola that has a vertex at
(
−
3
,
−
4
)
(-3,-4)
(
−
3
,
−
4
)
and the point
(
−
6
,
−
25
4
)
(-6,-\frac{25}{4})
(
−
6
,
−
4
25
)
that lies on the curve.
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Math Problems
Algebra 2
Write equations of parabolas in vertex form using properties
Full solution
Q.
Find the
a
a
a
value of a parabola that has a vertex at
(
−
3
,
−
4
)
(-3,-4)
(
−
3
,
−
4
)
and the point
(
−
6
,
−
25
4
)
(-6,-\frac{25}{4})
(
−
6
,
−
4
25
)
that lies on the curve.
Identify Vertex Form:
Identify the vertex form of a parabola.
\newline
Vertex form:
y
=
a
(
x
−
h
)
2
+
k
y = a(x-h)^2 + k
y
=
a
(
x
−
h
)
2
+
k
Plug in Vertex:
Plug in the vertex
(
−
3
,
−
4
)
(-3,-4)
(
−
3
,
−
4
)
into the vertex form.
\newline
y
=
a
(
x
+
3
)
2
−
4
y = a(x+3)^2 - 4
y
=
a
(
x
+
3
)
2
−
4
Substitute Point for 'a':
Substitute the point
(
−
6
,
−
25
4
)
(-6,-\frac{25}{4})
(
−
6
,
−
4
25
)
into the equation to find 'a'.
\newline
−
25
4
=
a
(
−
6
+
3
)
2
−
4
-\frac{25}{4} = a(-6+3)^2 - 4
−
4
25
=
a
(
−
6
+
3
)
2
−
4
Simplify Equation:
Simplify the equation.
\newline
−
25
4
=
a
(
−
3
)
2
−
4
-\frac{25}{4} = a(-3)^2 - 4
−
4
25
=
a
(
−
3
)
2
−
4
\newline
−
25
4
=
9
a
−
4
-\frac{25}{4} = 9a - 4
−
4
25
=
9
a
−
4
Move Terms:
Move
−
4
-4
−
4
to the left side of the equation.
\newline
−
25
4
+
4
=
9
a
-\frac{25}{4} + 4 = 9a
−
4
25
+
4
=
9
a
\newline
−
25
4
+
16
4
=
9
a
-\frac{25}{4} + \frac{16}{4} = 9a
−
4
25
+
4
16
=
9
a
Combine Fractions:
Combine the
fractions
on the left side.
\newline
(
−
25
+
16
)
/
4
=
9
a
(-25 + 16)/4 = 9a
(
−
25
+
16
)
/4
=
9
a
\newline
−
9
/
4
=
9
a
-9/4 = 9a
−
9/4
=
9
a
Divide to Solve for 'a':
Divide both sides by
9
9
9
to solve for 'a'.
\newline
−
9
4
÷
9
=
a
-\frac{9}{4} \div 9 = a
−
4
9
÷
9
=
a
\newline
−
9
4
×
1
9
=
a
-\frac{9}{4} \times \frac{1}{9} = a
−
4
9
×
9
1
=
a
Multiply Fractions:
Multiply the fractions to find
a
a
a
.
a
=
−
9
36
a = \frac{-9}{36}
a
=
36
−
9
a
=
−
1
4
a = \frac{-1}{4}
a
=
4
−
1
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\newline
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\newline
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\newline
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x
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y
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x
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x
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y
2
=
−
5
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\newline
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\newline
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\text{[A]up}
[A]up
\newline
[B]down
\text{[B]down}
[B]down
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(
_
,
_
)
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(
_
,
_
)
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Question
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y
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1
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(
y
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4
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\newline
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1
1
1
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\newline
(A) The center is
(
−
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0
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(
−
4
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1
1
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\newline
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(
4
,
0
)
(4,0)
(
4
,
0
)
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1
1
1
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\newline
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(
0
,
4
)
(0,4)
(
0
,
4
)
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1
1
1
.
\newline
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(
0
,
−
4
)
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(
0
,
−
4
)
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1
1
1
.
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Question
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x
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(
x
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y
=
(
x
−
3
)
(
x
+
9
)
\newline
The given equation is graphed in the
x
y
x y
x
y
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x
x
x
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\newline
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1
1
1
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\newline
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-3
−
3
and
−
9
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−
9
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9
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x
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5
x
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x
y
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y
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\newline
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y
y
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y
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