+0

0
266
3

a curve has equation y= 1/x^2 + 4x

a) find dy/dx

b) the point P(-1,-3) lies on the curve. Find an equation of the normal to the curve at the point P

c) find an equation of the tangent to the curve that is parallel to the line y= -12x

Nov 22, 2018

#1
+28468
+1

Nov 22, 2018
#2
0

What is the answer for C?

Guest Nov 22, 2018
#3
+28468
+3

c)  The tangent line parallel to y = -12x will have slope -12, so will be of the form y = -12x + k, where k is a constant.

We must have  -2x-3 + 4  = -12 for the slopes to match, which gives us the value of x at which this tangent line touches the curve.  Plug this value into y = x-2 + 4x to find the corresponding value of y.  Plug these values of x and y into  y = -12x + k and rearrange to find k.

-2x-3 + 4  = -12 → x-3 = 8  → x = 1/2

The corresponding value of y is (1/2)-2 + 4(1/2) → 4 + 2 → 6

So  6 = -12*(1/2) + k   →   6 = -6 + k  →  k = 12

Hence   y = -12x + 12 is the equation of the line tangent to the curve and parallel to y = -12x,

Alan  Nov 22, 2018