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

Jul 26, 2018

Below

Explanation:

$f \left(x\right) = \frac{2 x - 3}{{x}^{2} + x - 6}$

For vertical asymptotes, ${x}^{2} + x - 6 \ne 0$ as the denominator cannot equal to 0. If it equals to 0, then the function at the point is undefined. To find the vertical asymptotes, we find at what points the function is undefined

${x}^{2} + x - 6 = 0$
$\left(x + 3\right) \left(x - 2\right) = 0$
$x = - 3$ and $x = 2$

For horizontal asymptotes, $y = 0$ is the only asymptote.

Think of it this way:
When you sub random numbers into $x$, the denominator will always be bigger than the numerator. Hence, when you divide a small number by a larger number, then the answer will tend to 0 (be close to 0)

For intercepts,
When $y = 0$, $2 x - 3 = 0$, $x = \frac{3}{2}$
When $x = 0$, $y = \frac{1}{2}$

Plotting your intercepts and drawing in your asymptotes, you should get something like below. Remember, your asymptotes only affect/influence the endpoints of your graph and not anywhere else on your graph

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