Parabola Point Problem

First we note that the $y$ -intercept is $c$ . As the $x$ -intercept is $-5$ we see that $0 = (-5)^{2} + (-5)b + c \Longrightarrow c = 5b - 25$ .

Now as $dy/dx = 2x + b$ the slope of the tangent to the parabola at the $y$ -intercept is $b$ . Next, as the $y$ -intercept is the closest point on the parabola to $(12,3)$ the line joining the $y$ -intercept to this point will be perpendicular to the tangent line and thus have slope $-\dfrac{1}{b}$ . Thus

$-\dfrac{1}{b} = \dfrac{3 - c}{12 - 0} \Longrightarrow b = \dfrac{12}{c - 3}$ , which upon substitution into $c = 5b - 25$ gives us that

$c = \dfrac{60}{c - 3} - 25 \Longrightarrow (c + 25)(c - 3) = 60 \Longrightarrow c^{2} + 22c - 135 = 0 \Longrightarrow (c - 5)(c + 27) = 0$ .

So either $c = 5 \Longrightarrow b = 6$ or $c = -27 \Longrightarrow b = -\dfrac{2}{5}$ . But upon graphing these two results we note that only for the former is the $y$ -intercept the closest point to $(12,3)$ , so we can conclude that the parabola in question is $y = x^{2} + 6x + 5$ which has $y$ -intercept $\boxed{5}$ .