What is the equation of the tangent line of #f(x) =x^2+x+6# at #x=2#?

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Answer 1 · 2016-02-29T02:01:47

#y=5x+2#

Find the point the tangent line will intercept:

#f(2)=2^2+2+6=12rarr(2,12)#

Find the function's derivative through the power rule:

#f(x)=x^2+x+6#

#f'(x)=2x+1#

Find the slope of the tangent line at #x=2#:

#f'(2)=2(2)+1=5#

We will use point-slope form, which relates the point #(x_1,y_1)# and slope #m# into the equation:

#y-y_1=m(x-x_1)#

Here we have #(x_1,y_1)=(2,12)# and #m=5#, so the tangent line is

#y-12=5(x-2)#

Which simplifies to be

#y=5x+2#

Graph both the original function and tangent line to check:

graph{(x^2+x+6-y)(y-5x-2)=0 [-10, 10, -3.8, 28.68]}