Any (non-vertical) line through (9, 6.125) has equation of the form y= m(x- 9)+ 6.125.
You want to find m so that line is tangent to \(\displaystyle y= (-1/8)x^2+ 8\) and then determine "x" at the point where they are tangent.
If this were in the "Calculus" section, I would take the derivative of \(\displaystyle (-1/8)x^2+ 8\) and set it equal to m while setting \(\displaystyle m(x- 9)+ 6.125= (-1/8)x^2+ 8\), getting two equations to solve for m and x.
Since this is "Pre-Calculus", we can use a method, due to Fermat, that pre-dated Calculus (so, literally, "pre-Calculus"). Again we want x and m such that \(\displaystyle m(x- 9)+ 6.125= (-1/8)x^2+ 8\), since the line must touch the parabola at that point. That is, of course, a quadratic equation, \(\displaystyle (1/8)x^2+ mx- (9m+ 1.875)= (1/8)(x^2+ 9mx- (72m+ 15)= 0\).
Further, in order to be tangent, this x must be double root! The left side must be a perfect square, of the form \(\displaystyle (x- a)^2\). That, in turn, means that the discriminant, \(\displaystyle b^2- 4ac= (9m)^2+ 4(1)(72m+15)= 81m^2+ 288M+ 60= 0\).
Solve that equation for the two possible values of m.
