Question

What is the equation of the line tangent to `f(x)=9x^2 - 28x - 34` at `x=-1`?

Answer 1

`y=-64x+67`

Explanation:

`f(x)=9x^2-28x-34`

The tangent line is of the form `y=mx+c`. We can get `m` by taking the 1st derivative:

`f'(x)=18x-28`

So when `x=-1` then `m=(18xx-1)-28=-64`

To find the `y` coordinate:

`f(-1)=9+28-34=3`

Putting these values into the tangent equation to get `crArr`

`3=(-64xx-1)+c`

`:.c=67`

So the equation of the tangent line `rArr`

`y=-64x+67`

graph{(-64x+67-y)(9x^2-28x-34-y)=0 [-160, 160, -80, 80]}