H2 Maths 1.2 - Sketch a Rational Function with Stationary Points

Step 1 - The question

Sketch y equals x squared minus x plus four over x minus two, labelling all asymptotes, axial intercepts, and stationary points.

Step 2 - Vertical asymptote

The denominator is zero when x equals two, so x equals two is the vertical asymptote.

Step 3 - Oblique asymptote by long division

Divide x squared minus x plus four by x minus two to find the oblique asymptote.

Step 3 - Oblique asymptote by long division

x squared minus x plus four equals x minus two times x plus one, remainder six.

Step 3 - Oblique asymptote by long division

So y equals x plus one plus six over x minus two.

Step 3 - Oblique asymptote by long division

As x tends to plus or minus infinity, six over x minus two vanishes, and y approaches x plus one.

Step 3 - Oblique asymptote by long division

The oblique asymptote is y equals x plus one.

Step 4 - Intercepts

For the x-intercepts, check the discriminant of the numerator.

Step 4 - Intercepts

The discriminant is negative one squared minus four times four, which is one minus sixteen, equal to negative fifteen.

Step 4 - Intercepts

Since the discriminant is negative, the numerator has no real roots - there are no x-intercepts.

Step 4 - Intercepts

For the y-intercept, set x to zero: y equals four over negative two, which is negative two.

Step 5 - Stationary points

Differentiate using the rewritten form: y equals x plus one plus six over x minus two.

Step 5 - Stationary points

dy by dx equals one minus six over x minus two squared.

Step 5 - Stationary points

Set this to zero: x minus two squared equals six.

Step 5 - Stationary points

Taking square roots gives x minus two equals plus or minus root six.

Step 5 - Stationary points

So the stationary points are at x equals two plus root six and x equals two minus root six.

Step 6 - Classify and state coordinates

Substitute back to find the y-coordinates.

Step 6 - Classify and state coordinates

At x equals two plus root six: y equals two plus root six plus one plus six over root six, which simplifies to three plus two root six.

Step 6 - Classify and state coordinates

At x equals two minus root six: y equals three minus two root six.

Step 6 - Classify and state coordinates

The second derivative is positive at x equals two plus root six, so that is a local minimum.

Step 6 - Classify and state coordinates

The second derivative is negative at x equals two minus root six, so that is a local maximum.

Step 7 - Sketch and annotate

The curve has two branches separated by the vertical asymptote at x equals two.

Step 7 - Sketch and annotate

The right branch has a local minimum at two plus root six, three plus two root six.

Step 7 - Sketch and annotate

The left branch has a local maximum at two minus root six, three minus two root six.

Step 7 - Sketch and annotate

Both branches approach the oblique asymptote y equals x plus one as x tends to infinity.

Step 7 - Sketch and annotate

Common pitfall: the numerator has no real roots, so the curve does not cross the x-axis - do not draw x-intercepts.