Review question

# Can we sketch this function with its asymptotes? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5559

## Solution

Write down the equations of the two asymptotes of the curve $y = x+ 1 - \frac{4}{x-2}$ and find the coordinates of the points at which the curve meets the axes.

Sketch the curve.

When $x=2$, the denominator is zero, so there is a vertical asymptote at $x=2.$

If we look at what happens on either side of $x=2$, we see that $\dfrac{4}{x-2}$ gets very large and positive as $x$ approaches $2$ from the left, and $\dfrac{4}{x-2}$ gets very large and negative as $x$ approaches $2$ from the right.

As $x$ tends to infinity in either the positive or negative direction, the term $-\dfrac{4}{x-2}$ tends to $0$, so the graph of $y= x+ 1 - \dfrac{4}{x-2}$ approaches the line $y=x+1$. Therefore $y=x+1$ is the other asymptote.

Setting $x = 0$ gives $y = 3$, so the curve meets the $y$-axis at 3.

Setting $y = 0$ gives $x + 1 - \dfrac{4}{x-2} = 0$, which yields a quadratic equation. \begin{align*} & x + 1 - \frac{4}{x-2} = 0 \\ \iff & (x+1)(x-2) = 4 \\ \iff & x^2 - x - 6 = 0 \\ \iff & (x+2)(x-3) = 0 \\ \iff & x = 3 \textrm { or } x = -2. \end{align*}

So the curve meets the $x$-axis at $-2$ and $3$. We can now deduce the shape of the graph.