Need calc help!! find the tangent line to the curve y=x^3 at (1,1)?

Question

what do i do?? i tried using the formula f(a+h)-f(a)/h but i'm having trouble eliminating the h on the bottom.

Answer 1

y'=3x^2
at (1,1) y'=3
equation y-1=3(x-1)
3x-y=2 is the equation of the tangent at (1,1)

Answer 2

take the derivative of f(x) which is y=3x^2 then set x equal 1 to find the "b" of the equation so then you have the slope of the line that touches that point (1,1) and is parallel to the slope of f(x) at (1,1) but if you have not gotten into differentiation yet, you probably just have an alegebra problem, what you should find is that the a's should cancel and then you should be able to divide by h

yes basically (a+h)^3 -(a)^3 divided by h so just factor it out and the h will cancel on top

Answer 3

Take the derivative of the expression, i.e. dy/dx. In this case, the derivative is y' = 3x^2. At the point (1, 1), the derivative evaluates to 3, so all you need is a line with a slope of 3 that passes through (1, 1).

Answer 4

I haven't had calc in a while, but I believe you need the point-slope formula for that one. Find the slope of the curve at (1,1) and then put it into point-slope. You can find the slope at that point by figuring out the derivitive of y=x^3, then putting the point (1,1) into the derivitive.

Answer 5

Okay, I don't know if your class have done deriviatives, but that's how i would do it.
The derivative would give you the slope at that point.
The derivative is 3x^2.
SO at (1,1), it would be 3(1)^2 = 3.
So, put it in y=mx+b form, 1=(3*1)+b, so b = -2.
So, the equation is y=3x-2.

Answer 6

I've put a beautifully formatted and carefully annotated solution at the link below. Check it.
http://www.tomsmath.com/tangent-to-x-cubed-at-1-comma-1.html