How do you graph #y=-5/x-7# using asymptotes, intercepts, end behavior?

1 Answer
Jul 4, 2018

Answer:

See below:

Explanation:

We immediately notice that if our denominator is equal to zero, we'll be undefined and have a vertical asymptote at #x=0#.

Since we have a vertical asymptote, this means the graph never intercepts the #y#-axis since the function goes unbounded.

What about #x#-intercepts? We can easily find these by setting #y# equal to zero:

#-5/x-7=0#

Adding #7# to both sides gives us

#-5/x=7#

Let's multiply both sides by #7# to get

#-5=7x#

Dividing both sides by #7# gives us

#x=-5/7#

This is where our graph intercepts the #x#-axis. So already, we have a sense of how our graph looks, but end behavior can tell us more:

Since we know our function has a vertical asymptote, we know it goes unbounded towards infinity. What about negative infinity?

Let's evaluate the following limit:

#lim_(xto-oo)(-5/x-7)#

Since we will be dividing by a more and more negative number, #-5/x# will just go to zero, and we're left with #-7#. We can think of this as our horizontal asymptote.

Putting together all we know about our function, we can graph!

graph{-5/x-7 [-9.71, 10.29, -10.42, 0]}