# How do you graph y=(x-1)/(x-2)?

Aug 2, 2015

You must find first the discontinuity points and in order to graph it better you should find the OX and OY axis' cut points.

#### Explanation:

The most important thing in rational functions such as yours is to find the discontinuity points, which are those points that their image is infinity or do not share the same lateral limits. Let's explain it better:

You have a rational function, where both numerator and denominator are lineal functions, thus, the only way you can get a discontinuity in your graph is by having one point whose image is infinity, and this can only be got by those points whose image is shaped like $\frac{n}{0}$. So, our discontinuity points are those who cancel the denominator, those will be:

$x - 2 = 0 \to x = 2$

Now we need to know how the graph is going to diverge in that point, since its image is infinity, for this we must calculate his lateral limits.

${\lim}_{x \to {2}^{+}} \frac{x - 1}{x - 2} = \frac{1}{0} ^ + = + \infty$

Which means that for very very close values approaching to 2 by his right, their image tend to positive infinity, because a number slightly bigger than 2, minus 2, results a number slightly bigger than 0, that's called a positive 0, and since our numerator is a positive number either, the fraction results a positive infinite.
We reiterate the same process with numbers at his left.

${\lim}_{x \to {2}^{-}} \frac{x - 1}{x - 2} = \frac{1}{{0}^{-}} = - \infty$

So, we know the behavior of our function near 2, let's find out the function's inifite limits.

${\lim}_{x \to \infty} \frac{x - 1}{x - 2} = \frac{1}{1} = 1$

Taking the values with the highest degree, we see that for positive and negative huge values (tending to infinite) theis image tend to 1, hence we already know the vertical and horizontal asymptotes.

Now we should consider calculating the cut points of the function with the OX and OY axis'. Let's calculate the OX ones:

To know this, we must equal our image to 0 in order to move only horizontally:

$\frac{x - 1}{x - 2} = 0 \to x - 1 = 0 \to x = 1$

Hence our function only cuts the OX axis in x=1.
To find the cut point of the OY axis, we just need to calculate 0's image:

$f \left(0\right) = \frac{0 - 1}{0 - 2} = - \frac{1}{-} 2 = \frac{1}{2}$

And now with all this information, you can easily graph your function, which plots:
graph{(x-1)/(x-2) [-10, 10, -5, 5]}