Which values can $x(x-3)/(x-4)$ take?

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

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

Using the same axes, sketch the curve \(y=5/x\) 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.