Question

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

Answer 1

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 - 5x + 6`
`x^2 - 2x - 3x - 6`
`(x-2)(x-3)`

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

`(x-1)(x-3)`

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

`f(x) = [(x-2)(x-3)] / [(x-1)(x-3)]`

Now you are ready to graph with all of the criteria. Since `(x-3)` 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 `(x-1)` equal to zero. Then, solving for `x`, you get `x = 1`. The vertical asymptote of `f(x)` 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(x)` is `y = 1`.

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

You know the `y` - intercept is `(0, 2)` by substituting `x = 0` into the equation.

That's all. Hope that helps!

Answer 2

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 `(3, 1/2)` that would be in the graph.

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