Sketching Graphs C1

[br]And when $x$ is 0 you're left with $\frac{-1}{-3}$ which is the same as $\frac{1}{3}$.[br]

[br]'(b) Find the coordinates of the points where the curve with equation $y = f(x-1)$ crosses the coordinate axes.'[br]\vspace{0.2in}[br][br]The question is asking what the coordinates are when the curve crosses the $x$ and $y$-axis. When a function crosses the $x$-axis the $y$-value at that point must be 0, and when a function crosses the $y$-axis the $x$-value at that point must be 0.[br]\vspace{0.2in}[br][br]Assuming you got the answer right in part $(a)$, you could use that to find the point the curve crosses the $x$-axis. The original graph crosses at the origin (0,0), and $f(x-1)$ translated it so it crosses at (1,0).[br]\vspace{0.2in}[br][br]To find where it crosses the $y$-axis you put $f(x-1)=0$, so first you will have to change $f(x) = \frac{x}{x-2}$ (given in the question) into $f(x-1)$. To do this, every time you see an '$x$' in $f(x)$ you need to replace it with '$x-1$'. [br]\vspace{0.2in}[br][br]so $y = \frac{x-1}{(x-1)-2}$[br]= $\frac{x-1}{x-3}$[br]\vspace{0.2in}[br][br]And when $x$ is 0 you're left with $\frac{-1}{-3}$ which is the same as $\frac{1}{3}$.[br]