Question

Determine how many points the parabola has in common with the x-axis and whether it's vertex lies above, on, or below the x-axis.

Answer 1

Points "in common with the x-axis" are also known as the roots of the quadratic equation $x^2-12x+12=0$. You can apply the quadratic root formula to determine the roots, and also to determine how many such roots there are. With a quadratic (whose graph is a parabola), there can be maximum of 2 roots. But under certain circumstances, there may be only one or no such root.

The root formula for a generic quadratic $ax^2+bx+c$ is as follows:

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

The expression $b^2-4ac$ under the square root is called the determinant. It is called so because it determines the number of real roots. If the determinant value is > 0, there will be 2 roots (and so the parabola will cross the x-axis in 2 points), if its value is =0, there will be only a single root (the the parabola will touch the x-axis in exactly one point), and, finally, if its value is < 0, the quadratic has no real root (andthe parabola will not have any x-intercepts).

So, let's take a look:

$b^2-4ac= (-12)^2-4\cdot 1\cdot12=96$

This means the parabola will intercept the x-axis at 2 points, two real roots.

Since the coefficient of the quadratic term is positive (a=1), the parabola is oriented "open-up." But since we already know the parabola intercepts in two points, the fact that it is open-up implies now that the vertex must lie below the x-axis (otherwise it could not intercept it).