Question

find the equation of the tangent line to the graph f(x)=x^3 at the given x= -2

Answer 1

when x=-2
y=f(x)=(-2)^3=-8
so point is (-2,-8 )
next
first find the slope
sdlope is equal to derivative
differentiating
\[\Large m=\frac {dy}{dx}=3x^2\]
put x-2 so
\[\Large m=3(-2)^2=12\]

use point slope form
\[\Large y-y_{0}=m(x-x_{0})\]
where m=12 x0=-2 and y0=-8

put in the above equation to get the required equation of tangent line .

Answer 2

I got y=12x+16 as the answer.

Answer 3

this is correct .