Line Intersecting once with Parabola

Since this line passes through $(3,9)$ , then $3(m)+b=9 \Rightarrow b=9-3m\cdots(1)$

Also, this line is tangent to the graph of $y=x^2$ once so there is exactly one solution to the system

$\begin{cases} y=mx+b\cdots(2)\\ y=x^2\cdots(3)\\ \end{cases}$

Setting $(1)$ and $(2)$ equal, we get $y=mx+b=x^2 \Rightarrow x^2-mx-b$ has exactly one solution so the discriminant $(-m)^2-4(-b)=m^2+4b$ is $0$ . Substituting $(1)$ in, we have

$m^2+4(9-3m)=0 \Rightarrow m^2-12m+36=0 \Rightarrow (m-6)^2=0 \Rightarrow m=6 \Rightarrow b=9-3m=-9$

$m+b=6+(-9)=\boxed{-3}$ .

We proceed with a bit of calculus. Taking the derivative of $x^{2}$ at the point (3,9), we see that the slope is 6. To satisfy the other equation, $b$ must be -9. Therefore $6-9$ is -3. Hey Stephen! Enjoying Brilliant? Great problem!