Practice 1: calculus meeting algebra

Why this problem

Scholarship questions rarely stay inside one topic. This one starts as a curve-sketching question and turns into an algebra problem about a parameter — exactly the kind of join the examiners look for.

Work it yourself before reading the solution. Give it 20 minutes and write full sentences.

The problem

The curve $C$ has equation

$y = \frac{x^2 + k}{x - 1}, \qquad x \neq 1$

where $k$ is a constant.

(a) Show that $C$ has two stationary points when $k > -1$ , and none when $k < -1$ .

(b) Find, in terms of $k$ , the $x$ -coordinates of the stationary points.

(c) Describe what happens to the curve when $k = -1$ , and explain why.

Worked Example

Answer:

(a) We need $\dfrac{dy}{dx}$ , so apply the quotient rule with $u = x^2 + k$ and $v = x - 1$ :

$\frac{dy}{dx} = \frac{(2x)(x-1) - (x^2 + k)(1)}{(x-1)^2}$

Expanding the numerator:

$2x^2 - 2x - x^2 - k = x^2 - 2x - k$

$\frac{dy}{dx} = \frac{x^2 - 2x - k}{(x-1)^2}$

Stationary points occur where the numerator is zero (the denominator is never zero on the domain):

$x^2 - 2x - k = 0$

This is a quadratic in $x$ . The number of solutions is decided by the discriminant:

$\Delta = (-2)^2 - 4(1)(-k) = 4 + 4k = 4(1 + k)$

If $k > -1$ then $\Delta > 0$ , giving two distinct real roots — two stationary points.
If $k < -1$ then $\Delta < 0$ , giving no real roots — no stationary points.

This is the required result. Note the whole of part (a) is algebra about a discriminant; the calculus was only the first two lines.

(b) Solving $x^2 - 2x - k = 0$ by the quadratic formula:

$x = \frac{2 \pm \sqrt{4 + 4k}}{2} = 1 \pm \sqrt{1 + k}$

So the stationary points are at $x = 1 + \sqrt{1+k}$ and $x = 1 - \sqrt{1+k}$ .

Notice they are symmetric about $x = 1$ , which is the vertical asymptote — a structural fact worth stating.

(c) When $k = -1$ , $\Delta = 0$ , so the two stationary points coincide at $x = 1$ .

But $x = 1$ is excluded from the domain, so the stationary point does not exist on the curve. What has happened is that the numerator becomes $x^2 - 1 = (x-1)(x+1)$ , so

$y = \frac{x^2 - 1}{x - 1} = x + 1, \qquad x \neq 1$

The curve degenerates into a straight line with a single point removed at $x = 1$ . The asymptote has cancelled.

That cancellation is the insight the question is really testing.

Now write it up

Attempt part (c) in your own words before comparing. The mark is not for spotting the factorisation — it is for explaining why the asymptote disappears.

Test yourself

Write your own answer

Write a full answer first, then reveal the model answer and compare — that comparison is where the learning is.

In your own words, explain what happens to the curve $y = \frac{x^2+k}{x-1}$ when $k = -1$ , and why the vertical asymptote disappears. Write it as you would in the exam — full sentences, with the algebra that supports it.