What is the equation of the normal to the tangent line to #y = 8sqrt(x) - x# at #x = 1#?

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Answer 1 · 2017-09-27T13:37:13

The equation is #y = -1/6x + 43/6#

We have

#y(1) = 8sqrt(1) - 1/1 = 7#

The derivative is

#y' = 4x^(-1/2) + 2/x^3 = 4/sqrt(x) + 2/x^3#

The slope of the tangent is

#y'(1) = 4/sqrt(1) + 2/1#

#y'(1) = 4 + 2#

#y'(1) = 6#

So the slope of the normal will be #-1/6#, because the normal is always perpendicular to the tangent.

#y - y_1 = m(x - x_1)#

#y - 7 = -1/6(x - 1)#

#y - 7 = -1/6x + 1/6#

#y = -1/6x + 43/6#

Hopefully this helps!