How do you find the equation of the tangent line to the graph of #f(x)=x^3+1# at point (1,2)?
1 Answer
Sep 1, 2017
The derivative of
#f'(x) = 3x^(3 - 1) + 0(1)x^(0 - 1)#
#f'(x) = 3x^2#
We now determine the slope of the tangent line by plugging in the point
#f'(1) = 3(1)^2 = 3#
Now we can readily find the equation of the line.
#y -y_1 = m(x - x_1)#
#y - 2 = 3(x - 1)#
#y = 3x - 3 + 2#
#y = 3x - 1#
Now we can check the graphical interpretation and confirm that we are correct.
Hopefully this helps!