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

##### 1 Answer

#### Answer:

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 = -2/(x+3)-2 #

In the form:

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

We should now recognise this as the graph of the function

# y=1/x #

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

Which is:

1) **inverted** to give

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

2) **scaled** by a factor of

graph{-2/x [-10, 10, -8, 8]}

2) **translated** **left** to give

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

3) **translated** **down** to give

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

===========================================================

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

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

**End Behaviour:**

#x rarr +oo => x+3 rarr +oo => (-2)/(x+3) rarr 0^-#

#x rarr -oo => x+3 rarr -oo => (-2)/(x+3) rarr 0^+#

Hence the end behaviour is as follows:

# lim_(x rarr +oo) y = -2^- # and# lim_(x rarr -oo) y = -2^+ #

**Asymptotes**

The denominator is zero when

Hence, we have a vertical asymptote at

**Intercepts**

When:

# x=0 => y = 2/3-2 = -4/3 #

# y=0 => -1/(x+3)=1 => x=-4#

Hence, the intercepts are

Which is enough the sketch the curve.