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

Question

2319 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-06T18:12:11

This is a hyperbolic function with asymptotes $y = -1$ and $x = -2$

When $x = -2+epsilon$ for small $epsilon > 0$ , $y$ is large and positive.

When $x = -2-epsilon$ for small $epsilon > 0$ , $y$ is large and negative.

As $x->-oo$ we have $y->-1_-$

As $x->oo$ we have $y->-1_+$

The line of slope $1$ through the intersection $(-2, -1)$ of the asymptotes cuts the hyperbola at its two vertices $(-1, 0)$ and $(-3, -2)$

graph{1/(x+2)-1 [-12.08, 7.92, -5.68, 4.32]}

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