Answer: The asymptotes are \(x=-2\) and \(y=2x+1\).
\(\displaystyle \frac{\mathrm dy}{\mathrm dx}=2+3(x+2)^{-2}\gt 2\) for all \(x\neq -2\).
The curve has two increasing branches, one on each side of \(x=-2\), approaching the oblique asymptote \(y=2x+1\). It crosses the \(y\)-axis at \((0,-\frac12)\) and the \(x\)-axis at \(\left(\frac{-5\pm\sqrt{33}}4,0\right)\).
Divide \(2x^2+5x-1\) by \(x+2\):
\(2x^2+5x-1=(x+2)(2x+1)-3.\)
Hence
\(y=2x+1-\frac{3}{x+2}.\)
The denominator is zero at \(x=-2\), so the vertical asymptote is
\(x=-2.\)
As \(x\to\pm\infty\), the term \(-\dfrac{3}{x+2}\to0\), so the oblique asymptote is
\(y=2x+1.\)
Differentiate
\(y=2x+1-3(x+2)^{-1}.\)
Then
\(\frac{\mathrm dy}{\mathrm dx}=2+3(x+2)^{-2}.\)
Since \((x+2)^2\gt 0\) for every point of the curve, \((x+2)^{-2}\gt 0\). Therefore
\(\frac{\mathrm dy}{\mathrm dx}=2+3(x+2)^{-2}\gt 2.\)
For the sketch, \(y(0)=-\frac12\). The \(x\)-intercepts are obtained from \(2x^2+5x-1=0\), so
\(x=\frac{-5\pm\sqrt{25+8}}4=\frac{-5\pm\sqrt{33}}4.\)
The branch for \(x\gt -2\) lies below the oblique asymptote because \(-3/(x+2)\lt 0\), and the branch for \(x\lt -2\) lies above it because \(-3/(x+2)\gt 0\). Both branches are increasing.
