Question

What is the equation of the line tangent to `f(x)=x^2+6x-5` at `x=3`?

Answer 1

Equation of the line tangent is `12x-y-14=0`

Explanation:

At `x=3`, as `f(x)=x^2+6x-5=9+18-5=22`,

`f(3)=3^2+6xx3-5=9+18-5=22`.

Hence we are seeking a tangent at `(3,22)`.

Further slope of the tangent is given by the value of its first derivative at the point of tangent.

As first derivative `f'(x)=2x+6`, value of first derivative or slope of the desired tangent at `x=3` is `2xx3+6=12`.

As the tangent passes through `(3,22)` and has a slope of `12`,

equation of tangent is `(y-22)=12(x-3)` or `y-22=12x-36`

i.e.. `12x-y-14=0`
graph{(x^2+6x-5-y)(12x-y-14)=0 [-50, 50, -160, 160]}