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

Answer 1

Ezra Adams

#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)#.

Create a point-slope version of this.

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

Forming the slope-intercept:

#y=-12x-20#

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

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Answer 2

Scarlett Allison

The equation of the line tangent to f(x)=4x^2 + 4x - 4 at x=-2 is y = -8x - 4.

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Answer from HIX Tutor

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