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?
