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
Write the equation as y = x-2 + 4x.
a) Then dy/dx = -2x-3 + 4
b) The slope of the tangent line at P is thus (dy/dx)tangent = -2(-1)-3 + 4
The slope of the normal line at P is given by (dy/dx)normal = -1/(dy/dx)tangent .
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.
I'll leave you to do the actual calculations!