Question

What is the equation of the line tangent to `f(x)=x^2+8x - 12` at `x=3`?

Answer 1

`y=26x-57`

Explanation:

The gradient of the tangent at any point on a curve is given by the derivative of the function at that point.

So in this case, `f'(x)=2x^2+8`.

Therefore `f'(3)=2(3^2)+8`

`=26`.

But for the original function, `f(3)=3^2+8(3)-12=21`, and so the point `(3,21)` lies on the graph of `f`.

Furthermore, the tangent to `f` at `x=3` is a straight line and so must satisfy the equation of a straight line `y=mx+c`.

We may now substitute the point `(3,21)` as well as the gradient `26` of the tangent line into the general equation to solve for c, the y-intercept, and obtain :

`21=(26)(3)+c`, which implies that `c=-57`.

Thus the equation of the required tangent line is `y=26x-57`.