Question

How do you find the equation of the tangent line to the curve `f(x) = x^2 + 2x`; at x=3, x=5?

Answer 1

Tangent at `x=3` is `8x-y=9` and tangent at `x=5` is `12x-y=25`.

Explanation:

Slope of tangent at `x=x_0` on the curve `y=f(x)` is the value of `f'(x)` at `x=x_0` i.e. `x=x_0`.

Here we have `f(x)=x^2+2x`, hence `f'(x)=(df)/(dx)=2x+2`

Hence slope of tangent at `x=3` is `f'(3)=2*3+2=8`

and slope of tangent at `x=5` is `f'(5)=2*5+2=12`

Note that when `x=3`, `f(3)=3^2+2*3=15` hence tangent at `x=3` asses through `(3,15)` and has slope `8` and hence equation is `y-15=8(x-3)` i.e. `8x-y=9`.

and when `x=5`, `f(3)=5^2+2*5=35` hence tangent at `x=5` asses through `(5,35)` and has slope `12` and hence equation is `y-35=12(x-5)` i.e. `12x-y=25`.