Rational functions and asymptotes - TCE Mathematics Specialised (Tasmania)

Sketch $y = \dfrac{2x}{x - 1}$.

Step 1: Find the vertical asymptote

Set the denominator equal to zero to locate where the function is undefined. Each genuine zero of the denominator that is not cancelled by the numerator produces a vertical asymptote.

The denominator is zero at $x = 1$. The numerator $2x$ is not zero there, so there is a vertical asymptote at $x = 1$.

Step 2: Find the intercepts

Setting $x = 0$ gives the $y$-intercept, and setting the numerator to zero (excluding denominator zeros) gives any $x$-intercepts. Both intercepts are needed to anchor the sketch.

At $x = 0$: $y = \tfrac{0}{-1} = 0$, so the graph passes through the origin. This single point is both the $x$-intercept and the $y$-intercept.

Step 3: Determine the end behaviour

Compare the degrees of numerator and denominator to find the horizontal asymptote. When the degrees are equal, the asymptote is the ratio of the leading coefficients.

Both numerator and denominator have degree $1$, so the horizontal asymptote is:

Step 4: Analyse the sign and direction near each asymptote

Checking the sign of the function on each side of the vertical asymptote tells us whether each branch runs to $+\infty$ or $-\infty$, and knowing whether the curve approaches the horizontal asymptote from above or below completes the shape.

Just left of $x = 1$: the denominator is small and negative while the numerator is near $2$, so $y \to -\infty$. Just right of $x = 1$: the denominator is small and positive, so $y \to +\infty$. For large positive $x$, $y$ approaches $2$ from above; for large negative $x$, it approaches $2$ from below.

Final answer: the curve has two branches separated by the vertical asymptote $x = 1$, passes through the origin, and both branches approach the horizontal asymptote $y = 2$.

Sketch $y = \dfrac{x^2 + 1}{x}$.

Step 1: Find the vertical asymptote and intercepts

Set the denominator to zero to find the vertical asymptote. Then check for $x$-intercepts by solving $x^2 + 1 = 0$; since this has no real solutions ($x^2 + 1 > 0$ always), there are none. The function is also undefined at $x = 0$ so there is no $y$-intercept.

Vertical asymptote: $x = 0$. No $x$-intercepts and no $y$-intercept.

Step 2: Determine the end behaviour by polynomial division

When the numerator degree is exactly one more than the denominator degree, long division reveals an oblique asymptote. Dividing gives a linear quotient plus a remainder term that vanishes as $x \to \pm\infty$.

The oblique asymptote is $y = x$, and the remainder $\tfrac{1}{x}$ vanishes at both extremes.

Step 3: Locate turning points by differentiating

Writing the function in divided form makes differentiation straightforward. Setting the derivative to zero finds the local extrema, which determine the precise shape of each branch.

Differentiating $x + \tfrac{1}{x}$:

At $x = 1$: $y = 1 + 1 = 2$ (local minimum on the positive branch). By the odd symmetry of $x + \tfrac{1}{x}$, there is a local maximum at $(-1, -2)$ on the negative branch.

Step 4: Describe the final shape

For $x > 0$ the term $\tfrac{1}{x} > 0$ means the curve lies above the asymptote $y = x$; for $x < 0$ the term $\tfrac{1}{x} < 0$ means the curve lies below $y = x$. Combined with the turning points, this gives the full picture.

Final answer: the graph has a vertical asymptote at $x = 0$ and oblique asymptote $y = x$, with a local minimum at $(1, 2)$ and a local maximum at $(-1, -2)$. Both branches hug the line $y = x$ for large $|x|$.