1. ## graph the function...

graph the function. label all intercepts and asymptotes.

y = x-2 / (x+1)(x-3)

i got for the intercepts: y= 2/3
x= -2

y-asymptote = 2

how will this graph look?

2. Originally Posted by ohhhh
graph the function. label all intercepts and asymptotes.

y = x-2 / (x+1)(x-3)

i got for the intercepts: y= 2/3
x= -2

y-asymptote = 2

how will this graph look?
Check your y asymptote. I get x = -1 and x = 3 as asymptotes.

-Dan

3. I think you tried to solve x-2 (numerator) to find the horizontal asymptote. No, horizontal asymptotes are not found that way. You have to solve
$\lim_{\text{ }x\to \infty } \frac{x-2}{(x+1)(x-3)}$

4. yeah i did do the numerator.. woops. so, asymptotes at -1 and 3? my main concern is how this graph will look. i know it'll resemble the graph of 1/x .. what else do i need to do to figure our how the graph will look?

5. $\lim_{\text{ }x\to \mp \infty } \frac{x-2}{(x+1)(x-3)}$

$\lim_{\text{ }x\to \mp \infty } \frac{x-2}{x^2 - 2x -3}$

To calculate this limit, we divide both numerator and denominator by $x^n$, n is the highest power of x in the denominator. In $x^2 - 2x -3$, the highest power of x is 2. So n = 2. We'll divide both numerator and denominator by $x^2$.

$\lim_{\text{ }x\to \mp \infty } \frac{\frac{x-2}{x^2}}{\frac{x^2 - 2x -3}{x^2}}$

$\lim_{\text{ }x\to \mp \infty } \frac{\frac{1}{x}-\frac{2}{x^2}}{1 - \frac{2}{x} - \frac{3}{x^2}}$

When x goes to $\infty$, the fractions $\frac{1}{x}$, $\frac{1}{x^2}$, $\frac{2}{x}$, $\frac{3}{x^2}$... etc will approach 0.

So, the limit becomes $\frac{0}{1} = 0$. That means the horizontal asymptote is the line y=0, which is also called the x axis.