Is there any way to find range of a Quadratic/Quadratic function, without plotting its graph?
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$\begingroup$For $A\ne 0$ we have $Ax^2+Bx+C=Ag(x)+D$ where $g(x)=(x+\frac {B}{2A})^2$ and $D=C-\frac {B^2}{4A}.$ The range of $g(x)$ is $[0,\infty).$
If $A>0$ the range of $Ag(x)$ is $\{Ay: y\in [0,\infty)\}=[0,\infty).$ So the range of $Ax^2+Bx+C$ is $\{z+D: z\in [0,\infty)\}=[D,\infty).$
If $A<0$ the range of $Ag(x)$ is $\{Ay:y\in [0,\infty)\}=(-\infty,0].$ So the range of $Ax^2+BX+C$ is $\{z+D: z\in (-\infty,0]\}=(-\infty,D].$
$\endgroup$ $\begingroup$Because of the symmetry of the parabola, its stationary point will always be in the middle of its $x$-intercepts (or zeroes). Using the quadratic formula and taking the average of both roots, the $x$-coordinate of the stationary point of any quadratic function $ax^2+bx+c$ (where $a\neq0$) is given by $x=-\frac{b}{2a}$.
When $x=\frac{-b}{2a}$, $y=c-\frac{b^2}{4a}$.
Therefore the maximum or minimum value of the quadratic is $c-\frac{b^2}{4a}$. Whether it is the maximum or minimum can be determined by examining the sign of $a$.
If $a$ is positive, then the range is $y \geq c-\frac{b^2}{4a}$.
If $a$ is negative, then the range is $y \leq c-\frac{b^2}{4a}$.
$\endgroup$ $\begingroup$Hint:
The rational function $\dfrac{p(x)}{q(x)}$ has (one or two) vertical asymptotes if the denominator has real roots. It also has an horizontal asymptote $y=\dfrac{p_2}{q_2}$.
The local extrema are given by
$$\left(\frac{p(x)}{q(x)}\right)'=0$$ or
$$p'(x)q(x)=p(x)q'(x).$$
The two members are two cubic polynomial, with the same leading coefficient, so that this is in fact a quadratic equation (can also be linear or constant).
After solving for the roots, if any, the limits of the range are given by the value at the extrema, at the horizontal asymptote, and $\pm\infty$ if there are vertical asymptotes.
A longer analysis is required to exhaust the cases.
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