Question

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

Answer 1

`color(blue)(y=-9x+8)`

Explanation:

In order to find the equation of the tangent line, we must first find the gradient of this line at the given point.

To do this we find the derivative of `f(x)`, and then plug in `x=-1`

`f(x)=(x-3)^2-x`

We can expand this and then we only need to use the power rule to differentiate it.

`(x-3)^2-x=x^2-7x+9`

`dy/dx(x^2-7x+9)=2x-7`

Plug in `x=-1`

`2(-1)-7=-9`

This is our gradient `m=-9`

We need the corresponding `y` value when `x=-1`. Using `f(x)`

`y=((-1)-3)^2-(-1)=17`

Using point slope form of a line:

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

`y-17=-9(x-(-1))`

`color(blue)(y=-9x+8)`

GRAPH: