Parabolic X-Intercepts

This problem can be solved using surprisingly little calculus. First, construct the equation of the line tangent at x. To avoid confusion I will use x,y for the variables relating to the parabola equation, and X,Y for the variables of the line. Then, in point-slope form the line has equation $Y-(x^2+bx+c)=(2x+b)(X-x)$ . Rearranging this gives $Y=(2x+b)X-x^2+c$ . For the X intercept, set Y=0 and solve for X: $X=\frac{x^2-c}{2x+b}$ . Now, set that equal to the provided formula, so we have $\frac 12x-3=\frac{x^2-c}{2x+b}$ , or $x^2-c=x^2+(\frac 12b-6)-3b$ . These polynomials are equal for all x, so $\frac 12b-6=0$ and $c=3b$ . Solving these reveals $b=12,c=36$ , for a final answer of 48.

Since a line has an equation $y = y'x + y_0$ , where $y'$ is the slope and $y_0$ is the $y$ -intercept, the $y$ -intercept is $y_0 = y - y'x$ .

The $x$ -intercept $x_0$ is the $x$ value of $y = y'x + y_0$ when $y = 0$ , so $0 = y'x_0 + y_0$ which rearranges to $x_0 = \frac{-y_0}{y'}$ . Substituting $y_0 = y - y'x$ and simplifying gives $x_0 = x - \frac{y}{y'}$ .

Now we have $x_0 = x - \frac{y}{y'} = \frac{1}{2}x - 3$ , which rearranges to $\frac{y'}{y} = \frac{2}{x + 6}$ or $\frac{1}{y}dy = \frac{2}{x + 6} dx$ . Solving this differential equation gives $\ln |y| = 2 \ln |x + 6| + C$ for some constant $C$ . Letting $C = \ln k$ , this solves to $y = \pm k(x + 6)^2$ or $y = \pm k(x^2 + 12x + 36)$ .

Since the $x^2$ term must have a coefficient of $1$ , $\pm k = 1$ , and so $y = x^2 + 12x + 36$ , which means $b = 12$ and $c = 36$ , and $b + c = 12 + 36 = \boxed{48}$ .