Find the points where the tangent lines to the curve y = 2x^3 + 3x^2-12x + 1 are equal
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Answer:
Answer:
(-2,21) (1,6)
Explanation:
step one: find the derivative of the equation.
y
'
=
6
x
2
+
6
x
−
12
Step two: Since a horizontal line has a slope of 0, set the derivative to equal 0 and solve.
y
'
=
6
(
x
2
+
x
−
2
)
y
'
=
6
(
x
+
2
)
(
x
−
1
)
x
=
−
2
,
1
Step three: plug the x-values found in step 2 back into the original equation to get the y-coordinates of the points on the curve.
y
(
−
2
)
=
21
y
(
1
)
=
−
6
Step four: write out the coordinates of the points with a slope of zero.
(-2,21) and (1,-6)
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