Question

How do you find `a` for the derivative of `f(x)=x^2-ax` if the tangent has a y-intercept of -9 at the vertex?

Answer 1

`a=6`

Explanation:

Given the equation of the parabola:

`y= x^2-ax =x(x-a)`

we can immediately determine for symmetry reasons that the vertex has coordinates: `V = (a/2,-a^2/4)`, and the tangent line is:

Since the tangent to the parabola at the vertex is horizontal (because it is a local minimum) it follows that:

`-a^2/4 = -9`

That is:

`a=6`

graph{x^2-6x [-7.79, 12.21, -11.24, -1.24]}