Determine for which values of $\lambda$ the function

$$
y=\frac{x^{2}-6 x+5}{x^{2}+4 x+\lambda}
$$

has no extreme value. Examine the variation of this function and plot it if $\lambda=-12$.
A function does not have an extremum if it is constantly increasing or decreasing, which means its derivative is of constant sign and does not become zero. In the given case,

$$
\frac{d y}{d x}=\frac{(2 x-6)\left(x^{2}+4 x+\lambda\right)-(2 x+4)\left(x^{2}-6 x+5\right)}{\left(x^{2}+4 x+\lambda\right)^{2}}=\frac{10 x^{2}+2(\lambda-5) x-6 \lambda-20}{\left(x^{2}+4 x+\lambda\right)^{2}}
$$

The denominator of $\frac{d y}{d x}=y^{\prime}$ cannot be negative, so its sign is the same as the sign of the numerator. The numerator is a quadratic function of $x$; it is of constant sign and does not become zero if its discriminant is negative, which is the case if

$$
\begin{gathered}
D \equiv 4(\lambda-5)^{2}+40(6 \lambda+20) < 0
\end{gathered}
$$

Solving this inequality, we get $\lambda < -5$. If $\lambda < -5$, the function $y$ is constantly increasing.

$$
\text{The denominator of } y: x^{2}+4 x-12=(x+6)(x-2)
$$

This implies that the function $y$ is discontinuous at two points: $x=-6$ and $x=+2$. As $x$ increases, the value of the function becomes $+\infty$ at these points. Specifically, for a small positive $\varepsilon$:

$$
\begin{aligned}
& \lim y=+\infty, \text { if } x=-6-\varepsilon \text { and } \varepsilon \rightarrow 0 \\
& \lim y=-\infty, \text { if } x=-6+\varepsilon \quad " \quad " \\
& \lim y=-\infty, \text { if } x=2-\varepsilon \quad " \quad " \\
& \lim y=+\infty, \text { if } x=2+\varepsilon \quad " \quad " \text { " }
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_241379bd40f8fe87f3bdg-1.jpg?height=502&width=718&top_left_y=1454&top_left_x=690)

Thus, the curve of the function has three branches: the first between $-\infty$ and -6, the second between -6 and +2, and the third between +2 and $+\infty$. In the first branch, the function value increases from +1 to $+\infty$, in the second branch, it increases from $-\infty$ to $+\infty$, and in the third branch, it increases from $-\infty$ to +1. Indeed, if $x \rightarrow \pm \infty$, then $\lim y=+1$.

Accordingly, the line $y=1$ is an asymptote of the curve. The lines $x=-6$ and $x=+2$ are also asymptotes.