A function f is from R-------R such that f(x)=(a x^2+6x-8)/(a+6x-8x^2)
find the value of a for which for which f is onto.
$\displaystyle f(x)=y\Leftrightarrow \ldots \Leftrightarrow (a+8y)x^2+(6-6y)x-8-ay=0$
The above equation on $\displaystyle x$ has real solution iff:
$\displaystyle p_a(y)=\Delta=B^2-4AC=\ldots =(9+8a)y^2+(46+a^2)y+9+8a\geq 0$
We have a family of parabolas (a line if $\displaystyle a=-9/8$). We need to find $\displaystyle a$ such that $\displaystyle p_a(y)\geq 0$ for all $\displaystyle y\in\mathbb{R}$ . Necessarily $\displaystyle a>-9/8$ (why?) .
If $\displaystyle a>-9/8$ then, $\displaystyle f$ is onto iff $\displaystyle p(y_0)\geq 0$ being $\displaystyle y_0$ the "abscissa" of the $\displaystyle p_a$ vertex.
Let us see what do you obtain.