Let #f# be the function given by #f(x) = 2x^4-4x^2+1#. What is an equation of the line tangent to the graph at #(-2,17)#?

Let #f# be the function given by #f(x) = 2x^4-4x^2+1#. What is an equation of the line tangent to the graph at #(-2,17)#?

1 Answer
Oct 24, 2016

#y = -48x - 79#

Explanation:

The line tangent to the graph #y=f(x)# at a point #(x_0, f(x_0))# is the line with slope #f'(x_0)# and passing through #(x_0, f(x_0))#.

In this case, we are given #(x_0, f(x_0)) = (-2, 17)#. Thus, we only need to calculate #f'(x_0)# as the slope, and then plug that into the point-slope equation of a line.

Calculating the derivative of #f(x)#, we get

#f'(x) = 8x^3-8x#

#=> f'(-2) = 8(-2)^3-8(-2) = -64+16 = -48#

So, the tangent line has a slope of #-48# and passes through #(-2, 17)#. Thus, it's equation is

#y - 17 = -48(x - (-2))#

#=> y = -48x - 79#