Question

What is the equation of the tangent line of `f(x) =x^2+3x+2` at `x=-1`?

Answer 1

`color(blue)(y=x+1)`

Explanation:

To find the equation of the tangent line at `x=-1` we must first find the gradient of the line at the given point. If we differentiate `f(x)` this will give us a gradient function and allow us to find the gradient at any point on `f(x)`

Differentiating `f(x)=x^2+3x+2` using the Power Rule:

`dy/dx(ax^n)=anx^(n-1)`

Note that `dy/dx` is distributive over the sum:

i.e

`dy/dx(x^2+3x+2)=dy/dxx^2+dy/dx3x+dy/dx2`

`=2x+3+0=2x+3`

Plugging in `x=-1`

`2(-1)+3=1`

We now need a corresponding value of `y` where `x=-1`

Using `f(x)`

`(-1)^2+3(-1)+2=0`

Using point slope form of a line, where `m` is the gradient:

`(y_2-y_1)=m(x_2-x_1)`

`y-0=1(x-(-1))`

`color(blue)(y=x+1)`

GRAPH: