A Curve and Two Lines

Suppose the quadratic function is $g(x)=x^2+bx+c$ .

The two lines tangent to it can be rewritten as follows:

Line 1: $y=-5x+6$

Line 2: $y=x-1$

By definition of tangent, there must be $x_1,$ $x_2$ such that,

$x_1^2+bx_1+c=-5x_1+6$

$x_2^2+bx_2+c=x_2-1$

The two equations above can be expressed as follows:

$x_1^2+(b+5)x_1+(c-6)=0$

$x_2^2+(b-1)x_2+(c+1)=0$

By definition of tangent again, the discriminant should be zero:

$D_1=(b+5)^2-4(1)(c-6)=0$

$D_2=(b-1)^2-4(1)(c+1)=0$

Upon solving the system above, we get

$b=-\dfrac{13}{3}$

$c=\dfrac{55}{9}$ .

So, the sum of the coefficients of $g$ is given by

$1-\dfrac{13}{3}+\dfrac{55}{9}=\dfrac{25}{9}$ .

This leads to,

$p=25$

$q=9$ .

Hence,

$100p+q=100(25)+9=\boxed{2509}.$