Review question

# If we can sketch $y = f(x)$, can we sketch $y^2 = f(x)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6863

## Solution

Sketch the curve $y=\frac{(x-1)(x-4)}{(x-2)(x-3)}$

First, let’s think about the sign of $y$ for various values of $x$.

When $x=1$ or $x=4$, we have $y=0$. When $x=2$ or $x=3$, we see that $y$ is undefined—the graph has vertical asymptotes at these $x$-values.

For all other $x$-values, $y$ is either strictly positive or strictly negative. The sign of $y$ is determined by the sign of $x-1$, $x-2$, $x-3$ and $x-4$:

Range of x-values Sign of $y$
$x<1$ $y>0$
$1< x <2$ $y<0$
$2< x <3$ $y>0$
$3< x <4$ $y<0$
$4< x$ $y>0$
We can also think about what happens to $y$ as $x$ becomes large: we have \begin{align*} y &= \frac{(x-1)(x-4)}{(x-2)(x-3)} \\ &= \frac{x^2-5x+4}{x^2-5x+6} \\ &= 1-\frac{2}{x^2-5x+6}. \end{align*}

As $x \rightarrow \pm \infty$, the second term on the right tends to zero, and so $y$ tends to $1$.

We can look at the derivative of $y$ to see whether the graph has any stationary points. Using the quotient rule to differentiate the expression above for $y$, we have $\frac{dy}{dx}=\frac{2(2x-5)}{(x^2-5x+6)^2}.$

This is zero exactly when $x=\dfrac{5}{2}, y = 3$. As $(x^2-5x+6)^2$ is always non-negative, we can see that $\dfrac{dy}{dx}$ is negative for $x<\dfrac{5}{2}$, $x\neq 2$ and that $\dfrac{dy}{dx}$ is positive for $x>\dfrac{5}{2},$ $x\neq 3$.

The curve can be written as $y=\dfrac{\left(x-\frac{5}{2}\right)^2-\left(\frac{3}{2}\right)^2}{\left(x-\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2}$, and so is symmetrical about $x = \dfrac{5}{2}$.

We can now sketch the graph of this function.

…and also, using this curve if you wish and explaining your argument carefully, the curve $y^2=\frac{(x-1)(x-4)}{(x-2)(x-3)}.$

The variable $y$ is defined here only for values of $x$ for which $\dfrac{(x-1)(x-4)}{(x-2)(x-3)}$ is defined and non-negative, since $y^2$ must be positive.

Also note that if $\dfrac{(x-1)(x-4)}{(x-2)(x-3)}$ is defined and positive, the equation $y^2=\dfrac{(x-1)(x-4)}{(x-2)(x-3)}$ has two solutions in $y$—one the negative of the other, so the graph will be symmetrical about the $x$-axis.

We can also think about, for a particular $x$-value, whether the (positive) $y$-value at this point is larger or smaller in the second graph than in the first.

The positive $y$-value in the second graph is simply the square root of the $y$-value in the first.

So, if the $y$-value in the first graph is greater than $1$, it will be smaller at this point in the second graph.

If the $y$ value in the first graph is between $0$ and $1$, it will be greater in the second graph. If it is either $0$ or $1$, it is unchanged.

What about the gradient of this graph? We have $y^2=\frac{(x-1)(x-4)}{(x-2)(x-3)}.$

Differentiating implicitly with respect to $x$ gives us

$2y \frac{dy}{dx} = \frac{2(2x-5)}{(x^2-5x+6)^2}.$

What happens at $x=1$ and $x=4$ here? We have $y=0$, and so $\dfrac{dy}{dx}$ is undefined, which means that the graph is vertical at these $x$-values.

We can now sketch a graph of the function $y^2=\dfrac{(x-1)(x-4)}{(x-2)(x-3)}$.