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

1 Answer
Jul 9, 2016

Tangent line: #color(blue)(54x-y=90)#

Explanation:

Given
#color(white)("XXX")f(x)=x^3+4x^2+3#
at #x=3#
#color(white)("XXX")f(3)=3^3+4*3+3=27+36+9=72#
So tangent point we are interested in is at
#color(white)("XXX")#(3,72)#

The slope of the tangent for the general equation is given by the derivative of the function.
#color(white)("XXX")f'(x)=3x^2+8x+3#
when #x=3#
#color(white)("XXX")f'(3)=3*3^2+8*3+3=27+24+3=54#

So the tangent line at #x=3#
#color(white)("XXX")#has a slope of #54#
#color(white)("XXX")#and passes through the point #(3,72)#

Using the slope-point form for the tangent line:
#color(white)("XXX")y-72=54(x-3) = 54x-162#
or
#color(white)("XXX")y=54x-90#
or
#color(white)("XXX")54x-y=90#