Thread: Find the equation of line tangent

Originally Posted by Latszer

Find the equation of the line tangent to the graph of f at (2, 7), where f is given by f(x) = 2x^3 - 9x^2 + 3.

I'm confused. Step by step if at all possible.

Equation of tangent: y - 7 = m(x - 2) where m = f'(x) evaluated at x = 2.

Originally Posted by Latszer

Find the equation of the line tangent to the graph of f at (2, 7), where f is given by f(x) = 2x^3 - 9x^2 + 3.

I'm confused. Step by step if at all possible.

1) Find the derivative of f(x)
f(x)=2x^3-9x^2+3
f'(x)=6x^2-18x

2) Plug in x (2) to the derivative to find the slope.
m=6(2)^2-18(2)
m=6(4)-36
m=24-36
m=-12

3) Use y-y1=m(x-x1) to find the slope of the tangent line
y=7, x=2, m=-12
y-7=-12(x-2) You can finish here, or distribute if you want.