What is the equation of the line tangent to # f(x)=x^4-3x^2+2 # at # x=0#?

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Answer 1 · 2015-12-20T06:07:05

#y=2#

The line will be tangent to the point #(0,f(0))#.

#f(0)=2#, so the tangent line will pass through the point #(0,2)#.

The slope of the tangent line can be found through finding #f'(0)#.

#f'(x)=4x^3-6x#

#f'(0)=0#

Thus, since the slope of the tangent line is #0#, it will be the horizontal line #y=2#.

This also means that the point #(0,2)# is a relative minimum or maximum, since #f'(0)=0#.

We can check on a graph of #f(x)#:
graph{x^4-3x^2+2 [-10, 10, -5, 5]}