How do you find the equation of a line tangent to a graph #f(x)=2x^2# at (3,1)?
1 Answer
Sep 11, 2015
In the pint slope form it would be y-1=12(x-3)
Explanation:
Find the slope first, f'(x) = 4x. At (3,1) slope would be 4(3)=12.
Equation of the line tangent at (3,1) in the point slope form would be y-1=12(x-3)
