Question

1. determine the simplified linear function for $f(x)=\frac{2x^2 - 15x + 7}{x - 7}$ and algebraically determine the coordinates of the point(s) of discontinuity. 2 marks

Explanation:

Step1: Find where denominator is zero

A function is discontinuous where its denominator is zero (for rational functions). So, set \(x - 7=0\).
\(x - 7 = 0\implies x = 7\).

Step2: Check numerator at \(x = 7\)

Now, check the numerator \(2x^{2}-15x + 7\) at \(x = 7\).
\(2(7)^{2}-15(7)+7=2\times49-105 + 7=98-105 + 7=0\). Wait, but maybe we can factor the numerator. Let's factor \(2x^{2}-15x + 7\).
\(2x^{2}-15x + 7=2x^{2}-14x - x + 7=2x(x - 7)-(x - 7)=(2x - 1)(x - 7)\).

Step3: Simplify the function

Now, \(f(x)=\frac{(2x - 1)(x - 7)}{x - 7}\), for \(x
eq7\), we can cancel \(x - 7\). So \(f(x)=2x - 1\) for \(x
eq7\). But at \(x = 7\), the original function is undefined (denominator zero), but the simplified function \(2x - 1\) is defined at \(x = 7\) (value \(2(7)-1 = 13\)). So the discontinuity at \(x = 7\) is a removable discontinuity (a hole), and the point of discontinuity is \((7,13)\) (since the limit as \(x\to7\) is \(13\), and the function is undefined at \(x = 7\) in the original, but the simplified function would have that point).

Answer:

The point of discontinuity is \((7, 13)\)