- #1
danizh
- 16
- 0
#1: 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.
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.