Double the value of a quadratic expression If I have this expression $f(x)=ax^2+bx+c$ and I want an expression for $2f(x)$ can I get that with a given change of $x$ ? For example, I have $f(x)=2x^2+3x+9$ and I want to find which factor for $x$ would make $f(x)$ twice the value at every point. Basically I'm looking for another function $g(x)$ so that I can use that and then $f(g(x))=2f(x)$.
 A: For some functions $f$, there exists no such function $g$ (assuming that you’re working in the real numbers only).
Consider for example $f(x) = 1 - x^2$. We have $f(0)= 1$, but there exists no value of $x$ for which $f(x) = 2f(0)= 2$.
A: Let's try your example, $f(x)=2x^2+3x+9$.
You require that
$$ 2(g(x))^2+3g(x)+9 = f(g(x))= 2f(x)=4x^2+6x+18.$$
As a quadratic equation in $g(x),$ this is
$$ 2(g(x))^2 + 3g(x) - (4x^2+6x+9) = 0$$
and its solution by the quadratic formula is
\begin{align}
g(x) &= \frac{-3 \pm \sqrt{3^2 - 4(2)(- (4x^2+6x+9))}}{2(2)} \\
&= \frac{-3 \pm \sqrt{32x^2 + 48x + 81}}{4} \\
\end{align}
Pick your choice of either $+$ or $-$ for the $\pm$ sign and you have a function,
for example,
$$
g(x) = -\frac34 + \frac14 \sqrt{32x^2 + 48x + 81}.
$$

Note that $f(x)=2x^2+3x+9$ has no real roots.
For a quadratic polynomial with just one real root,
$g(x)$ can be simpler.
For example, if we let $f(x) = x^2 + 4x + 4,$ then
$$ (g(x))^2+4g(x)+4 = f(g(x)) = 2f(x)=2x^2+8x+8, $$
$$ (g(x))^2+4g(x)-(2x^2+8x+4) = 0, $$
\begin{align}
g(x) &= \frac{-4 \pm \sqrt{4^2 - 4(- (2x^2+8x+4))}}{2} \\
&= -2 \pm \frac14 \sqrt{8x^2 + 32x + 32} \\
&= -2 \pm \frac14 \sqrt{8(x^2 + 4x + 4)} \\
&= -2 \pm \frac{\sqrt2}{2} (x + 2) \\
\end{align}
and you can choose a sign from $\pm$ to get a single function such as
$$ g(x) = -2 + \frac{\sqrt2}{2} (x + 2) 
=  \frac{\sqrt2}{2} x + \left({\sqrt2} - 1\right). $$

If the quadratic polynomial has two roots, however, it will have either a minimum value that is negative or a maximum value that is positive.
If there is a minimum value, it is not possible to get twice the minimum value by substituting $g(x)$ for $x$ in the polynomial, because twice the minimum is less than the minimum (which is negative) and by definition there is nothing you can plug into the polynomial that will give you less than the minimum.
For similar reasons, if there is a maximum you cannot get twice the maximum.
The best you can do is to double the polynomial's value for values of $x$ outside some interval around the value where the minimum or maximum of the polynomial occurs.
A: For which values of $x$ this is possible will depend on the quadratic polynomial:
You want to determine, for each $x$, whether one can find an $h$ such that $f(x+h)=2f(x)$.
Remember that Taylor's formula is an exact formula for polynomials. For a quadratic polynomial, this yields
$$f(x+h)=f(x)+ f'(x)h+\frac{f''(x)}2 h^2,$$
so that you have to solve for
$$f'(x)h+\frac{f''(x)}2 h^2 =f(x)\iff  f''(x)  h^2+2f'(x)h-2f(x)=0$$
which is a quadratic equation in $h$, with reduced discriminant \begin{align}
\Delta'(x)&=f'(x)^2+2f(x)f''(x)=(2ax+b)^2+4a(ax^2+bx+c)\\
&=8a^2x^2+6abx+b^2+2ac,
\end{align}
a quadratic polynomial in $x$. There remains to consider the (reduced) discriminant of $\Delta'(x)$  – $a^2(b^2-8ac)$ and consider the  cases $\:b^2-8ac >0, <0, =0$.
