For what values of x, if any, does #f(x) = 1/((5x+8)(x-6) # have vertical asymptotes?

1 Answer
mason m
Jan 29, 2016

#x=-8/5# and #x=6#

Explanation:

For a rational function such as the one presented, vertical asymptotes occur whenever the denominator of the function is equal to #0#. These are the "holes" in the function's domain since it's impossible to divide by #0#.

Now, we must find the times when

#(5x+8)(x-6)=0#

What we have here are two terms, #5x+8# and #x-6#, that are being multiplied and equalling #0#. This means that either one of the terms must be equal to #0#, but it doesn't matter which.

Thus, to find the times when the whole expression equals #0#, we can set either term equal to #0#.

#mathbf((1))#

#5x+8=0#

#5x=-8#

#x=-8/5#

There is a vertical asymptote at #x=-8/5#.

#mathbf((2))#

#x-6=0#

#x=6#

There is a vertical asymptote at #x=6#.

We can check a graph of the original function:

graph{1/((5x+8)(x-6)) [-7.75, 12.25, -4.915, 5.085]}

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