What is the equation of the line tangent to #f(x)= 2x^3+8x # at #x=-1#?

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Answer 1 · 2016-01-06T19:13:18

#y+24=14(x+1)#, or, if you prefer slope-intercept form, the equation is #y=14x-10#

#f(x) = 2x^3+8x#

At #x= -1#, we have #y=f(-1) = 2(-8)+8(-1) = -16-8 = -24#.

So the tangent line contains the point #(-1,-24)#

#f'(x) = 6x^2+8#, so the slope of the tangent at the point where #x= -1# is

#m = f'(-1) = 6(1)+8 = 14#

The line through #(-1, -24)# with slope #m=14# has equation:

#y+24=14(x+1)#.

If you prefer slope-intercept form, the equation is #y=14x-10#