Question

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

Answer 1

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`