Solution

Find the range of values of \(p\) for which the graph of \(y = px^2 + 8x + p - 6\) crosses the \(x\)-axis.

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 \(-2<p<8\).

State also the values of \(p\) for which the \(x\)-axis is a tangent to the curve.

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 \(p = -2\).