## Problem 2

Determine the values of the real parameter $a$, such that the function $f: D \rightarrow \mathbb{R}, f(x)=\frac{x^{2}-5 x+4}{x^{2}+a^{2} x+2 a}$ has a single vertical asymptote.
## Problem 2

The function $f: D \rightarrow \mathbb{R}$ has a single vertical asymptote $\Leftrightarrow \mathbf{1}$) the equation $x^{2}+a^{2} x+2 a=0$ has a unique solution or 2) the equations $x^{2}+a^{2} x+2 a=0$ and $x^{2}-5 x+4=0$ have a unique common real root

1) $\Delta=0 \Leftrightarrow a^{4}-8 a=0 \Leftrightarrow a\left(a^{3}-8\right)=0 \Rightarrow a_{1}=0, a_{2}=2$

For $a=0 \Rightarrow D=\mathbb{R} \backslash\{0\}, f(x)=\frac{x^{2}-5 x+4}{x^{2}} \Rightarrow \lim _{x \rightarrow 0} f(x)=\infty \Rightarrow x=0$ is the unique vertical asymptote.

For $a=2 \Rightarrow D=\mathbb{R} \backslash\{-2\}, f(x)=\frac{x^{2}-5 x+4}{(x+2)^{2}} \Rightarrow \lim _{x \rightarrow-2} f(x)=\infty \Rightarrow x=-2$ is the unique vertical asymptote.....1p
2) $x^{2}-5 x+4=0 \Rightarrow x_{1}=1, x_{2}=4$. If $x_{1}=1$ is the common root $\Rightarrow 1+a^{2}+2 a=0 \Rightarrow(a+1)^{2}=0 \Rightarrow a=-1$

$\Rightarrow f: \mathbb{R} \backslash\{-2,1\} \rightarrow \mathbb{R}, f(x)=\frac{x^{2}-5 x+4}{x^{2}+x-2}=\frac{x-4}{x+2} \Rightarrow \lim _{\substack{x \rightarrow-2 \\ x-2}} f(x)=\infty$

$\Rightarrow x=-2$ is the unique vertical asymptote.

If $x_{2}=4$ is the common root $\Rightarrow 16+4 a^{2}+2 a=0 \Rightarrow \Delta<0 \Rightarrow a \in \varnothing \Rightarrow a \in\{0,2,-1\}$ is the solution.