Answer:

Write

\(\frac{2x^2-x+5}{x^2-1}=2+\frac{A}{x-1}+\frac{B}{x+1}.\)

Then

\(2+\frac{A}{x-1}+\frac{B}{x+1}=\frac{2(x^2-1)+A(x+1)+B(x-1)}{x^2-1}.\)

The numerator is

\(2x^2+(A+B)x+(A-B-2).\)

Equating coefficients with \(2x^2-x+5\) gives

\(A+B=-1,\qquad A-B=7.\)

So \(A=3\) and \(B=-4\), hence

\(\frac{2x^2-x+5}{x^2-1}=2+\frac{3}{x-1}-\frac{4}{x+1}.\)

Differentiate this form:

\(\frac{dy}{dx}=-\frac{3}{(x-1)^2}+\frac{4}{(x+1)^2}.\)

For stationary points, set \(\frac{dy}{dx}=0\):

\(\frac{4}{(x+1)^2}=\frac{3}{(x-1)^2}.\)

So

\(4(x-1)^2=3(x+1)^2,\)

which simplifies to

\(x^2-14x+1=0.\)

The discriminant is

\((-14)^2-4(1)(1)=192\gt 0,\)

so there are two distinct real values of \(x\) for which \(\frac{dy}{dx}=0\).

The vertical asymptotes are \(x=1\) and \(x=-1\), and the horizontal asymptote is \(y=2\).

At \(x=0\),

\(y=\frac{5}{-1}=-5,\)

so the curve crosses the \(y\)-axis at \((0,-5)\).

To find where the curve meets \(y=2\), solve

\(\frac{2x^2-x+5}{x^2-1}=2.\)

This gives \(2x^2-x+5=2x^2-2\), so \(x=7\). Hence the point is \((7,2)\).

For \(x\)-axis intersections, set \(y=0\):

\(2x^2-x+5=0.\)

The discriminant is \(1-40=-39\lt 0\), so there are no real \(x\)-intercepts.

The sketch should therefore show three branches with asymptotes \(x=\pm1\) and \(y=2\), crossing the \(y\)-axis at \((0,-5)\), meeting \(y=2\) at \((7,2)\), and no \(x\)-axis intersections.