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?