What is the equation of the line tangent to # f(x)=4x^2 + 4x - 4# at # x=-2#?

Question

6971 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-20T15:31:14

#y=-12x-20#

The tangent line will pass through the point #(-2,f(-2))#.

#f(-2)=4(-2)^2+4(-2)-4=4#

Thus, the tangent line will pass through the point #(-2,4)# and have a slope of #f'(-2)#.

#f'(x)=8x+4#

#f'(-2)=8(-2)+4=-12#

The slope of the tangent line is #-12# and it passes through #(-2,4)#.

Write this in point-slope form.

#y-4=-12(x+2)#

In slope-intercept form:

#y=-12x-20#

graph{(y+12x+20)(4x^2+4x-4-y)=0 [-23.14, 17.41, -8.89, 11.37]}