1) We can find the slant asymptotes by working with limits. So let's begin with that, by calculating the side limits of this function to check its behavior.
[tex]\begin{gathered} \lim _{x\to-\infty\: }\frac{f\left(x\right)}{x}\: \Rightarrow\lim _{x\to-\infty\: }\frac{\sqrt{25+36x^2}}{x}=\frac{\sqrt{25+36x^2}}{x} \\ \lim _{x\to\: -\infty\: }\mleft(-\sqrt{\frac{25}{x^2}+36}\mright) \\ -\sqrt{\lim_{x\to\:-\infty\:}\left(\frac{25}{x^2}+36\right)} \\ -\sqrt{\lim_{x\to\:-\infty\:}\left(\frac{25}{x^2}\right)+\lim_{x\to\:-\infty\:}\left(36\right)} \\ \lim _{x\to\: -\infty\: }\mleft(\frac{25}{x^2}\mright)=0 \\ \lim _{x\to\: -\infty\: }\mleft(36\mright)=36 \\ \mathrm{Simplify\: }-\sqrt{0+36}=-6 \\ y=-6x \\ y=6x\mathrm{\: }\mleft(\mathrm{slant}\mright),\: y=-6x\mathrm{\: }\mleft(\mathrm{slant}\mright) \end{gathered}[/tex]
Note that here we performed some limits properties, a bit simplified due to the time. But the point here is the behavior of this function.
2) As we know the slant asymptotes are y=6x and y=-6x so we cn
