Find the tangent y x 3 2x 4 where the curves y=x^3-3x+4
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Tangent:
y
=
2
x
−
11
Normal:
y
=
−
1
2
x
−
6
Explanation:
A tangent line would have a gradient equal to that of the curve at x=2, while a normal line would have a gradient which is the negative reciprocal of that of the curve at x=2
d
y
d
x
=
3
x
2
−
4
x
−
2
When x=2,
d
y
d
x
=
3
(
2
)
2
−
4
(
2
)
−
2
=
2
Hence, tangent at x=2:
y
=
(
d
y
d
x
)
x
+
c
=
2
x
+
c
Solve for c by setting
x
=
2
,
y
=
(
2
)
3
−
2
(
2
)
2
−
2
(
2
)
−
3
=
−
7
Thus,
c
=
−
7
−
2
(
2
)
=
−
11
The equation of the tangent line is thus
y
=
2
x
−
11
graph{(y-x^3+2x^2+2x+3)(y-2x+11)=0 [1, 3, -7.5, -6.5]}
Similarly, for the normal line,
y
=
−
1
2
x
+
c
c
=
−
7
−
(
−
1
2
(
2
)
)
=
−
6
Thus the equation of the normal line is
y
=
−
1
2
x
−
6
graph{(y-x^3+2x^2+2x+3)(y+1/2x+6)=0 [1, 3, -7.5, -6.5]}
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