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

Guest Nov 21, 2018

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!

Alan  Nov 21, 2018

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