Q:

Find the asymptote of $y=\frac{x^2+x+7}{\sqrt{2x+1}}$

Find the asymptote of $y=\frac{x^2+x+7}{\sqrt{2x+1}}$

I found one of the asymptotes which is $x=-\frac{1}{2}$ but couldn't find the other. I tried using L'Hopital's rule but it turns out the answer I got was wrong after I plotted the graphs. And then I'm stuck. Can anyone help me with this? Thank you guys.

A:

If we factor out $\frac1{\sqrt2}$ from the denominator and complete the square in the numerator, we get
$$y=\frac1{\sqrt2}\cdot\frac{(x+1/2)^2+27/4}{\sqrt{x+1/2}}=\frac1{\sqrt2}\left((x+1/2)^{3/2}+\frac{27/4}{\sqrt{x+1/2}}\right)$$
The other asymptote may be found by dropping the last term:
$$y\sim\frac1{\sqrt2}(x+1/2)^{3/2}$$
This is a curvilinear asymptote.

