How do you graph #y=-2/(x-7)# using asymptotes, intercepts, end behavior?

1 Answer
Mar 16, 2017

See explanaition


First you should find the vertical asymptotes of the equation. This is done by setting the denominator equal to zero.
so #x=7# is the vertical asymptote.

Then since the numerator's highest exponent on a variable is lower than the denominator, the horizontal asymptote is #y=0#

Then you can plug in a point left of the vertical asymptote like 6 for the x value. This will output 2 for the y value. we know it is above the horizontal asymptote so it must be increasing because it cannot cross either the vertical or horizontal asymptote.

You can draw a line that starts horizontally towards the left side and curves upward as it reaches the asymptote (which it never touches).

in other words

#lim_(x rarr -oo) f(x)=0#


#lim_ (x rarr 7^-) f(x)=oo#

Now you can repeat this same process on the right side of the asymptote.

Plug in a point on the right side of the asymptote for x
you will get a negative value. So you can figure out:

#lim_ (x rarr oo) f(x)=0#

#lim_ (x rarr 7^+) f(x)=-oo#

graph{-2/(x-7) [-3.88, 16.12, -4, 6]}