#1:(adsbygoogle = window.adsbygoogle || []).push({}); The tangent at a point P on the curve y = x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

So far, I found the derivative of y = x^3, which is y' = 3x^2. Now I set these two equal to each other:

X^3 = 3x^2

x^3 - 3x^2 = 0

x^2 (x-3) = 0

so x = 0 or x = 3.

I don't know where to go from here, or if I did this right at all.

#2:Show that the x- and y- intercepts for any tangent to the curve y = 16 - 8sqrt(x) + x have a sum of 16.

First I expanded the equation:

y = 16 - 8sqrt(x) + x

= 16 - 8^(1/2) + x

Then I determined the derivative:

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

= -4/sqrt(x) + 1

Once again, I have no idea what do from here. :(

Any help would be greatly appreciated.

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# Homework Help: Two Calculus Problem Solving Questions.

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