Doubly Tangent Parabola

We have

$Q(x) = a x^2 + b x + c$

as the quadratic function whose graph is tangent to $f(x)$ at two distinct points. Define

$g(x) = f(x) - Q(x)$

Then, since $f(x)$ opens upward, it follows that, $f(x)$ has to be above $Q(x)$ for all $x$ except at the points of tangency, where they are coincident. This means $g(x)$ is always non-negative.

Now,

$g(x) = x^4 - 8 x^3 + (17-a) x^2 + (2 - b) x + (-24 - c)$

Since $g(x)$ is non-negative and has two roots, then these two roots must be double roots, and therefore,

$g(x) = ( x^2 + e x + f )^2$

for specific constants $e, f$ .

Expanding,

$g(x) = x^4 + 2 e x^3 + (2f + e^2 ) x^2 + 2 e f x + f^2$

Comparing the two forms of $g(x)$ , we deduce that

$e = -4$ , and

$17 - a = e^2 + 2 f = 16 + 2 f \implies a = 1- 2f$ , and

$2 - b = 2 e f \implies 2 - b = -8 f \implies b = 2 + 8 f$

and $f^2 = -24 - c \implies c = -24 - f^2$

We also have, (from the fact that the quadratic passes through $(2, -20)$ that

$4 a + 2 b + c = -20$

Combining all four equations, we get

$4(1- 2f) + 2(2 + 8f) + (-24 - f^2) = -20$

$-16 +8f- f^2 = -20$

$f^2 - 8 f + 16 = 20$

$(f - 4)^2 = 20$

$f = 4 \pm \sqrt{20}$

The discriminant condition reveals that we must have

$f = 4 - \sqrt{20}$

for two real roots of $g(x)$ to exist.

Hence, $a = 1.9443$ , $b = -1.77709$ , $c = -24.2229$

It follows that

$| a | + | b | + | c | = 27.944$