Find the slopes of the tangent lines to the graph of x^3 at the point (1,1) , (2,8) , and (3,27).?

Question

My calc teacher is horrible and explains things so bad. I really don't understand how to do this problem. Thank you!

Answer 1

Find the derivative (3x^2) and plug in the different x-values you have. Don't worry about the y-values.
3(1)^2 = 3
3(2)^2 = 12
3(3)^2 = 27

The derivative is the instantaneous rate of change, or the slope of the tangent, at any point on the graph. It gives you a formula usually to find that slope. The formula for these problems in the beginning are in terms of x, so you just plug in the x you want to use and the value you get back is the slope at that point.

Answer 2

f (x) = x ³
f `(x) = 3 x ²
f `(1) = 3 = m1 , slope at (1,1)
f `(2) = 12 = m2 , slope at (2,8)
f `(3) = 27 = m3 , slope at (3 , 27)