f(x)=x2−4x+9(x+5)(x−3)
Let g(x)=x2−4x+9D<0g(x)>0 for x∈R 
$\therefore\left[\begin{array}{l}
\mathrm{f}(-5)=0 \
\mathrm{f}(3)=0
\end{array}\right.$
So, f(x) is many-one. again, $\begin{aligned}
& y x^2-4 x y+9 y=x^2+2 x-15 \
& x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 \
& \text { for } \forall x \in R \Rightarrow D \geq 0 \
& D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 \
& 5 y^2+2 y+16 \leq 0 \
& (5 y-8)(y+2) \leq 0
\end{aligned}$ 
