Find the parabola with equation y = ax^2 + bx whose tangent line at (2, 4) has equation y = 8x − 12.

[tex]\bf y=ax^2+bx\implies \cfrac{dy}{dx}=2ax+b[/tex]

now, we know the tangent line at (2,4) is y = 8x - 12, now, that's already in slope-intercept form, so, the slope of that tangent at (2,4) is "8" then.

now, that means when x = 2, the slope is 8, so let's use that.

[tex]\bf \left. \cfrac{dy}{dx} \right|_{x=2}\implies 2a(2)+b=\stackrel{\textit{from tangent equation}}{8}\implies 4a+b=8 \\\\\\ 4a=8-b\implies a=\cfrac{8-b}{4}\implies \boxed{a=2-\cfrac{b}{4}}\\\\ -------------------------------\\\\ \begin{cases} x=2\\ y=4\\\\ a=2-\cfrac{b}{4} \end{cases}\implies \stackrel{y = ax^2+bx}{4=\left( \boxed{2-\frac{b}{4}} \right)(2)^2+b(2)}\implies 4=8-b+2b \\\\\\ \boxed{-4=b}\qquad thus\qquad a=2-\cfrac{-4}{4}\implies \boxed{a=3}\\\\ -------------------------------\\\\ y=3x^2-4x[/tex]