Where does $y = px^2 + 8x + p - 6$ cross the $x$-axis?

Where does $y = px^2 + 8x + p - 6$ cross the $x$-axis?

Question

Find the range of values of $p$ for which the graph of $y = px^2 + 8x + p - 6$ crosses the $x$-axis. State also the values of $p$ for which the $x$-axis is a tangent to the curve.

Solution

Given a quadratic equation $ax^2 + bx + c = 0$, its discriminant is $d = b^2 - 4ac$.

The graph of $y = ax^2 + bx + c$ crosses the $x$-axis if, and only if, $d > 0$.

Here the discriminant is

$$\begin{align*} d &= 8^2 - 4 \times p \times (p - 6) \\ &= 64 - 4(p^2 - 6p) \\ &= 4(-p^2 + 6p + 16) \\ &= 4(8-p)(2+p). \end{align*}$$

The graph of $d=4(8-p)(2+p)=4(-p^2+6p+16)$ is “vertex-up” (as the coefficient of $p^2$ is negative), and is zero when $p=-2$ or $p=8$. Therefore $4(8-p)(2+p)>0$ if and only if $-2 < p < 8$.

Thus the graph crosses the $x$-axis if and only if $-2<p<8$.

This happens precisely when the $x$-axis touches the parabola. This happens if, and only if, the discriminant is zero.

From the above, this is the case when $p = 8$ or $p = -2$.