# How do you graph f(x)=(x^2-5x+6)/(x^2-4x+3) using holes, vertical and horizontal asymptotes, x and y intercepts?

Apr 24, 2017

See below.

#### Explanation:

It would be very hard to graph this equation correctly as without factoring both the numerator and denominator it would be hard to account for all of the information.

${x}^{2} - 5 x + 6$
${x}^{2} - 2 x - 3 x - 6$
$\left(x - 2\right) \left(x - 3\right)$

Then, following the same process for the bottom, you get

$\left(x - 1\right) \left(x - 3\right)$

Based on this, you can rewrite f(x) as:

$f \left(x\right) = \frac{\left(x - 2\right) \left(x - 3\right)}{\left(x - 1\right) \left(x - 3\right)}$

Now you are ready to graph with all of the criteria. Since $\left(x - 3\right)$ cancels out on both top and bottom, you know that there must be a hole at $x = 3$. To find the $y$-coordinate of the hole, simply substitute $x = 3$ into the factored equation, and the result is the $y$ - coordinate. Mark the hole's coordinates with a small, open circle.

Vertical asympotes can be found in the denominator, just set $\left(x - 1\right)$ equal to zero. Then, solving for $x$, you get $x = 1$. The vertical asymptote of $f \left(x\right)$ is $x = 1$.

For the horizontal asymptote, in this case, you would take the ratio of the leading coefficients of the numerator and denominator. Since this is $1$, then the horizontal asymptote of $f \left(x\right)$ is $y = 1$.

Zeros are found in the top. You set $\left(x - 2\right)$ equal to zero, and the result is $x = 2$. Therfore, there must be a "zero" or x-intercept at $\left(2 , 0\right)$.

You know the $y$ - intercept is $\left(0 , 2\right)$ by substituting $x = 0$ into the equation.

That's all. Hope that helps!

Apr 24, 2017

See graph

#### Explanation:

In addition to to the answer below, here is what the graph would look like. Keep in mind however, that this graph show the "hole" at $\left(3 , \frac{1}{2}\right)$ that would be in the graph.

graph{(x^2-5x+6)/(x^2-4x+3) [-10, 10, -5, 5]}