Question

find the two -intercepts of the function and show that f'(x) = 0 at some point between the two x -intercepts.

Answer 1

When we are given a quadratic function of the form:

$f(x)=ax^2+bx+c$

We can solve for:

$ax^2+bx+c=0$

Using the quadratic formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Since we want to find the x-intercepts, these are the x such that f(x) = 0. The function given in the problem is: