Thread: Find points on the graph...

Q: Find all points on the graph of y = (1/3)X^3 + X^2 - X where the tangent line has slope 1
im not sure how to solve this problem or where to start. can some one help?

Originally Posted by viet

Q: Find all points on the graph of y = (1/3)X^3 + X^2 - X where the tangent line has slope 1
im not sure how to solve this problem or where to start. can some one help?

Points on the graph of:

y = (1/3)x^3 + x^2 - x

where the tangen has slope 1, are the points where dy/dx=1.

So you need to differentiate (1/3)x^3 + x^2 - x with respect to x, and then
set the derivative equal to zero and solve the resulting equation. This will
give the x values where the slope is 1, the y values are obtained by
substituting these x's back into y = (1/3)x^3 + x^2 - x.

RonL

ok so the derivative is x^2+2x-1, x = -2, x = +sqrt(8), x = -sqrt(8)
then plug x into y = (1/3)X^3 + X^2 - X ?

Originally Posted by viet

ok so the derivative is x^2+2x-1, x = -2, x = +sqrt(8), x = -sqrt(8)
then plug x into y = (1/3)X^3 + X^2 - X ?

The derivative is dy/dx=x^2+2x-1, this equals one so:

x^2+2x-1=1,

so we need the roots of:

x^2+2x-2=0

which are x= -1+/- sqrt(3), these are plugged into y=y = (1/3)x^3 + x^2 - x
to find the corresponding values of y.

RonL

ahh ok, thank you so much CaptainBlack.

-tom