How do you find the vertex in y=−2(x+3)(x−1)?
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The vertex will have an
x
coordinate that is the average of the two zeros, which are at
x
=
−
3
and
x
=
1
.
So the
x
coordinate is at
x
=
−
3
+
1
2
=
−
1
Substitute this value of
x
back into the equation to get
y
=
−
2
(
x
+
3
)
(
x
−
1
)
=
−
2
(
−
1
+
3
)
(
−
1
−
1
)
=
−
2
×
2
×
−
2
=
8
So the vertex is at (-1, 8).
Another way of calculating this is as follows:
y
=
−
2
(
x
+
3
)
(
x
−
1
)
=
−
2
(
x
2
+
2
x
−
3
)
=
−
2
x
2
−
4
x
+
6
Differentiate this by
x
...
d
d
x
(
−
2
x
2
−
4
x
+
6
)
=
−
4
x
−
4
The derivative, which represents the slope of the curve at any point will be zero at the vertex, when
−
4
x
−
4
=
0
, that is when
x
=
−
1
.
Substitute this value of
x
back into the equation as before to get
y
=
8
, giving the vertex as (-1, 8).
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