Answer: Asymptotes
The denominator is zero when \(x=2\), so there is a vertical asymptote \(x=2\).
Divide \(x^2+px+1\) by \(x-2\):
\(x^2+px+1=(x-2)(x+p+2)+(2p+5)\).
So
\(y=x+p+2+\frac{2p+5}{x-2}\).
Hence the oblique asymptote is \(y=x+p+2\).
Distinct turning points
Differentiate:
\(y'=1-\frac{2p+5}{(x-2)^2}=\frac{(x-2)^2-(2p+5)}{(x-2)^2}=\frac{x^2-4x-(2p+1)}{(x-2)^2}.\)
Turning points occur when \(y'=0\), so
\(x^2-4x-(2p+1)=0\).
For two distinct turning points this quadratic must have two distinct real roots, so its discriminant is positive:
\((-4)^2-4(1)(-(2p+1))\gt 0\)
\(16+8p+4\gt 0\)
\(20+8p\gt 0\)
\(p\gt -\frac52\).
Sketch when \(p=-1\)
Then
\(y=\frac{x^2-x+1}{x-2}=x+1+\frac{3}{x-2}.\)
The asymptotes are \(x=2\) and \(y=x+1\).
Intercepts:
- \(y\)-intercept: at \(x=0\), \(y=\frac{1}{-2}=-\frac12\), so the curve passes through \((0,-\tfrac12)\).
- \(x\)-intercepts: solve \(x^2-x+1=0\). Its discriminant is \(1-4=-3\lt 0\), so there are no real \(x\)-intercepts.
So the sketch should show a vertical asymptote at \(x=2\), an oblique asymptote \(y=x+1\), the point \((0,-\tfrac12)\), and no crossings of the \(x\)-axis.
Let
\(y=\frac{x^2+px+1}{x-2}.\)
Find the asymptotes
The denominator is zero at \(x=2\), so the curve has a vertical asymptote
\(x=2.\)
Now divide \(x^2+px+1\) by \(x-2\):
\(x^2+px+1=(x-2)(x+p+2)+(2p+5).\)
Therefore
\(y=x+p+2+\frac{2p+5}{x-2}.\)
As \(x\to\pm\infty\), the fraction tends to \(0\), so the oblique asymptote is
\(y=x+p+2.\)
Values of \(p\) for two distinct turning points
Differentiate the rewritten form:
\(y'=1-(2p+5)(x-2)^{-2}=1-\frac{2p+5}{(x-2)^2}.\)
Putting over a common denominator gives
\(y'=\frac{(x-2)^2-(2p+5)}{(x-2)^2}=\frac{x^2-4x+4-2p-5}{(x-2)^2}=\frac{x^2-4x-(2p+1)}{(x-2)^2}.\)
Turning points occur when \(y'=0\), so we need
\(x^2-4x-(2p+1)=0.\)
This must have two distinct real roots, so its discriminant must be positive:
\(\Delta = (-4)^2-4(1)(-(2p+1))=16+8p+4=20+8p.\)
Hence
\(20+8p\gt 0 \implies p\gt -\frac52.\)
So the curve has two distinct turning points exactly when \(p\gt -\frac52\).
Sketch for \(p=-1\)
Substitute \(p=-1\):
\(y=\frac{x^2-x+1}{x-2}=x+1+\frac{3}{x-2}.\)
So the asymptotes are
\(x=2\quad\text{and}\quad y=x+1.\)
For the intercepts:
- At \(x=0\), \(y=\frac{1}{-2}=-\frac12\), so the curve crosses the \(y\)-axis at \((0,-\frac12)\).
- For \(x\)-intercepts, solve \(x^2-x+1=0\). The discriminant is \(1-4=-3\lt 0\), so there are no real \(x\)-intercepts.
So the sketch should show a vertical asymptote at \(x=2\), an oblique asymptote \(y=x+1\), the point \((0,-\frac12)\), and no intersection with the \(x\)-axis.
