find the points on the curve y= x^3-3x^2-4x at which the tangent is parallel to lines 4x+y-3=0
Answers
Step-by-step explanation:
The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. (If needed, then the normal is perpendicular to the tangent so the product of their gradients is
−
1
).
We have:
y
=
x
3
+
5
x
2
Then differentiating wrt
x
, gives us:
d
y
d
x
=
3
x
2
+
10
x
Now comparing the the line equation
y
=
11
x
−
π
to that of a line in standard form,
y
=
m
x
+
c
, we see immediately that
m
=
11
So any tangent to our curve that is parallel must have this same gradient. This will occur if and only if:
d
y
d
x
=
11
∴
3
x
2
+
10
x
=
11
∴
3
x
2
+
10
x
−
11
=
0
We can solve this quadratic equation using the quadratic formula, giving:
x
=
−
10
±
√
10
2
−
4
⋅
3
⋅
(
−
11
)
2
⋅
3
=
−
10
±
√
100
+
132
6
=
−
10
±
√
232
6
=
−
10
±
2
√
58
6
=
−
5
3
±
1
3
√
58
=
−
4.2053
,
0.8719
(4dp)
Using the curve equation we can find the full coordinates:
x
=
−
4.2053
⇒
y
=
14.0544
x
=
0.8719
⇒
y
=
4.6441
So there are two coordinates on the curve:
(
−
4.2053
,
14.0544
)
and
(
0.8719
,
4.6441
)
PLEASE MARK ME BRILLIANT