Question

Find an equation of the line tangent to the graph of f(x) at the point (−1, 7): f(x) =x^2+ 2x+ 8

Answer 1

derivative of function gives slope so df(x)/dx is 2x+2 at given point slope is 0 so it is parallel to x axis then equation is y=7

Answer 2

Answer:

The equation of tangent is y=7.

Step-by-step explanation:

The given function is

$f(x)=x^2+2x+8$

$f'(x)=2x+2$

The slope of tangent is

$f'(x)_{(-1,7)}=2(-1)+2=0$

The slope of tangent at point (-1,7) is 0. The equation of tangent is

$y-y_1=m(x-x_1)$

$y-7=0(x-(-1))$

$y=7$

Therefore the equation of tangent is y=7.