Thread: finding equation of tangent line

finding equation of tangent line at point (1,6)
y=(x²+2x)(x+1)

i got the derivative to be
f'(x)= 3x²+6x+2

but how do i use this to find the equation of a line at point (1,6)?
(im guessing i could use the y-y=slope(x-x) forumla but im unsure how with this derivative...)

-thx

you are correct

y = m(x-x0) + y0

m = f '(1) (x0,y0) = (1,6)

ans would be this?

11x-5?

i didnt quite understand your reply lol

that's correct

thx