# How do you graph y=-2/(x+3)-2 using asymptotes, intercepts, end behavior?

Sep 14, 2017

graph{-2/(x+3)-2 [-10, 10, -8, 8]}

#### Explanation:

I would advocate that just because we have more advanced mathematical skills , it does not mean that we should automatically use them in favour of basic translation skills.

We can write the given equation:

$y = - \frac{2}{x + 3} - 2$

In the form:

$\left(y + 2\right) = - \frac{2}{\left(x + 3\right)}$

We should now recognise this as the graph of the function

$y = \frac{1}{x}$

graph{1/x [-10, 10, -8, 8]}

Which is:

1) inverted to give $y = - \frac{1}{x}$
graph{-1/x [-10, 10, -8, 8]}

2) scaled by a factor of $2$ to give $y = - \frac{2}{x}$
graph{-2/x [-10, 10, -8, 8]}

2) translated $3$ units to the left to give $y = - \frac{2}{x + 3}$
graph{-2/(x+3) [-10, 10, -8, 8]}

3) translated $2$ units down to give $\left(y + 2\right) = - \frac{2}{x + 3}$
graph{-2/(x+3)-2 [-10, 10, -8, 8]}

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In the spirit of the question if we must use asymptotes, intercepts, end behavior then:

$y = - \frac{2}{x + 3} - 2$

End Behaviour:

$x \rightarrow + \infty \implies x + 3 \rightarrow + \infty \implies \frac{- 2}{x + 3} \rightarrow {0}^{-}$
$x \rightarrow - \infty \implies x + 3 \rightarrow - \infty \implies \frac{- 2}{x + 3} \rightarrow {0}^{+}$

Hence the end behaviour is as follows:

${\lim}_{x \rightarrow + \infty} y = - {2}^{-}$ and ${\lim}_{x \rightarrow - \infty} y = - {2}^{+}$

Asymptotes

The denominator is zero when $x + 3 = 0$

Hence, we have a vertical asymptote at $x = - 3$

Intercepts

When:

$x = 0 \implies y = \frac{2}{3} - 2 = - \frac{4}{3}$
$y = 0 \implies - \frac{1}{x + 3} = 1 \implies x = - 4$

Hence, the intercepts are $\left(0 , - \frac{4}{3}\right)$ and $\left(- 4 , 0\right)$

Which is enough the sketch the curve.