Can we sketch the curve $y=(x-a)e^x/(x-b)$?

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Let’s find the stationary points of \[y=\left(\frac{x-a}{x-b}\right)e^x.\]

We plan to differentiate using the product rule here, so it helps us to find the derivative of \(\dfrac{x-a}{x-b}\) first.

If \(z = \dfrac{x-a}{x-b}\), then by the quotient rule, \(z' = \dfrac{a-b}{(x-b)^2}\). Returning to our original problem, we have \(y'=0\) when \[\frac{dy}{dx}=\left(\frac{x-a}{x-b}\right)e^x+\left[\dfrac{a-b}{(x-b)^2}\right]e^x=0\] \[\implies \frac{e^x(x^2-(a+b)x+(a+ab-b))}{(x-b)^2}=0.\]

We know that \(e^x > 0\) for any \(x\), so we need \(x^2-(a+b)x+(a+ab-b)=0.\)

This has two solutions as long as the discriminant is positive, which happens if and only if \((a+b)^2 -4(a+ab-b) > 0.\)

Multiplying out and rearranging here gives \[(a-b)(a-b-4)>0,\] which holds when \[a-b<0 \quad \text{ or } \quad a-b>4.\]

The vertical asymptote to the curve is at \(x=\dfrac{1}{2}\). Now let’s substitute in the values of \(a\) and \(b\) to find the stationary points: \[x^2-(a+b)x+ab-b+a=x^2-\frac{1}{2}x-\frac{1}{2}=0.\] Solving this, we find that \[\begin{align*} &x^2-\frac{1}{2}x-\frac{1}{2}=0 \\ &\implies 2x^2-x-1=0\\ &\implies (2x+1)(x-1)=0\\ &\implies x = 1, -\dfrac{1}{2}, &\end{align*}\]

which lie on either side of the asymptote \(x=\dfrac{1}{2}\).

We can now sketch the graph. We know two stationary points, \(\left(-\dfrac{1}{2},\dfrac{e^{-1/2}}{2}\right)\) and \((1,2e)\), and by inspection we can see that it also passes through \((0,0)\).

We also have that \(y\to\infty\) as \(x\to+\infty\), and \(y\to 0\) as \(x\to -\infty\).

Also, as \(x\to \dfrac{1}{2}\) from the right, \(y\to+\infty\) and as \(x \to \dfrac{1}{2}\) from the left \(y \to -\infty\).

We may now give an approximate sketch of the graph.

We again find the stationary points of the curve, this time solving \[\begin{align*} &x^2-\frac{9}{2}x+\frac{9}{2}=0 \\ &\implies 2x^2-9x+9=0\\ &\implies (2x-3)(x-3)=0\\ &\implies x = 3, \dfrac{3}{2}. \end{align*}\]

So our two stationary points are \(\left(\frac{3}{2},-2e^{3/2}\right)\) and \(\left(3,-2e^3\right)\).

The vertical asymptote is at \(x=0\). As \(x\to 0\) from the right, we have that \(y\to-\infty\), since in this case \(x-\tfrac{9}{2} < 0\) and the denominator \(x\) is positive, so \(y\) is negative. As \(x\to 0\) from the left, we have \(y\to +\infty\) as \(x-\tfrac{9}{2} < 0\) again, but this time the denominator \(x\) is negative.

The graph crosses the \(x\)-axis once, at \(x=\dfrac{9}{2}\), and as \(x \to +\infty\), \(y \to +\infty\) and as \(x \to -\infty\), \(y\to 0\).