Step-by-step explanation:
Your graph will have a vertical asymptote at x = 1, as the denominator x - 1 cannot equal zero. Therefore, f(1) is not posssible. Then, plugging in test values for the function on the left side of the asymptote...
[tex]f(-1) = \frac{2(-1)-1}{-1-1} = \frac{3}{2}\\f(0) = \frac{2(0)-1}{0-1} = 1\\f(2) = \frac{2(2)-1}{2-1} = 3\\f(3) = \frac{2(3)-1}{3-1} = \frac{5}{2}[/tex]
... you will find that your graph needs to include the following points:
[tex](-1, \frac{3}{2})\\(0, 1)\\(2, 3)\\(3, \frac{5}{2})[/tex]
So therefore, your graph must cross through the above points, and may never cross the vertical line x = 1.
