Even after looking up how to do this and reading about it I still don't know how to find the slope of curve or the deritives . Please help solve these problems.

1. Y =x^3 p(-1, -1)
2. Y = x^4 /2 p(-1 , 1/2)
3. Y= 8x^4 -7x^2 +5x + 6 p(-1,2)

1 answer

if you don't know that the slope of the curve is given by the derivative, I suggest more study, or google will provide many discussions.

Assuming that you believe that the derivative of a function gives the slope of the curve at any point, then you just have to remember your Algebra I and the point-slope form of a line. In these problems, you work out the slope, and they give you the point.

SO, for the first one,
y = x^3
y' = 3x^2
That means that at any point (x,x^3), the tangent line has slope 3x^2. So, at (-1,-1), the tangent line has slope 3.

Now you have a slope and a point, so the tangent line has equation

y+1 = 3(x+1)
or,
y = 3x+2

See the line is tangent to the curve at

http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E3,+y%3D3x%2B2

follow these steps on the others. Use wolframalpha.com to verify your work.
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