How do you find the slope and the equation of the tangent line to the graph of the function at the given value of x for #f(x)=x^4-13x^2+36# at x=2?

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Answer 1 · 2015-09-03T22:03:53

The slope is #-20#, the equation is #y=-20x+40#.

First of all, the slope is given by the derivative, so you have that

#f'(x) = 4x^3 - 26 x#, and thus #f'(2)= 32-52=-20#

Also, we know that the tangent line must pass across the point #2,f(2)#, namely #(2, 0)#

If you know that a line has slope #m#, and crosses the point #(x_0,y_0)#, then you obtain its equation with the formula

#y-y_0 = m(x-x_0)#. Plugging the values, we have

#y=-20(x-2)#, which we can rearrange into #y=-20x+40#

As you can see in this link , the line is actually tangent in the requested point.