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.
On
If by asymptote you mean a line, then there's no other asymptote.
Why? Because
$$f(x)=\frac{x^2+x+7}{\sqrt{2x+1}}$$
which can be rewritten as:
$$f(x)=\frac{1}{4}\cdot (2x+1)\sqrt{2x+1} + \frac{1}{4}\cdot\frac{27}{\sqrt{2x+1}}$$
So there does not exist a line
$$g(x) = ax + b$$ such that:
$$\lim_{x \to \infty} |f(x)-g(x)| = 0$$
Why?
Because this term $(2x+1)\sqrt{2x+1}$ grows faster than any linear function as $x$ goes to infinity.
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.