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?

1 Answer
Jul 12, 2018

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]}