Mathematics - Functions Question with Solution | TestHub

$f(x)=\frac{(x+5)(x-3)}{x^2-4x+9}$
$\begin{array}{rl}& \text{ Let }g(x)=x^2-4x+9\\ & D<0\\ & g(x)>0\text{ for }x\in R\end{array}$
$\therefore [\begin{array}{l}\mathrm{f}(-5)=0\\ \mathrm{f}(3)=0\end{array}$
So, $\mathrm{f}(\mathrm{x})$ is many-one. again, $\begin{array}{rl}& y{x}^2-4xy+9y=x^2+2x-15\\ & x^2(y-1)-2x(2y+1)+(9y+15)=0\\ & \text{ for }\forall x\in R\Rightarrow D\ge 0\\ & D=4(2y+1{)}^2-4(y-1)(9y+15)\ge 0\\ & 5y^2+2y+16\le 0\\ & (5y-8)(y+2)\le 0\end{array}$