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
This was answered here: https://web2.0calc.com/questions/a-curve-has-equation-y-1-x-2-4x#r1
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,