Review question

# Which values can $x(x-3)/(x-4)$ take? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9391

## Suggestion

If $y=\frac{x(x-3)}{x-4}$ and $x$ is real show that $y$ cannot take any value between $+1$ and $+9$.

Could we write $y=\frac{x(x-3)}{x-4}$ as a quadratic in $x$?

Can we now use the discriminant of this?

Or alternatively, maybe calculus might help us here?

Can we find the stationary points?

Sketch the curve $y=\dfrac{x(x-3)}{x-4}$.

Here are some thoughts to consider while sketching any graph:

• Where does the graph intersect the axes?

• Where is the graph positive (above the $x$-axis) and where is it negative?

• Does the graph have any asymptotes, horizontal, vertical or diagonal?

• How does the graph behave as $x\to\pm\infty$?

• Does the graph have any stationary points? If so, where, and what is their nature (maximum, minimum, point of inflection)?

Using the same axes, sketch the curve $y=\dfrac{5}{x}$ and find its intersection with the line $x=4$.

Hence prove that the equation $\frac{5}{x}=\frac{x(x-3)}{x-4}$ has no positive root.

What is going on in the two separate regions $x<4$ and $x>4$ here?