What's the domain of #g(x)=(x^2-5x+4)/(x^2-2x-15)#?

1 Answer

Answer:

#x!=5,-3#

Explanation:

We have the function:

#g(x)=(x^2-5x+4)/(x^2-2x-15)#

Let's first talk about the values of #x# that won't be allowed in the domain. Why would an #x# value be disallowed?

In this case, when dealing with a fraction, we can't have the denominator be 0. So what values of #x# will make the denominator 0?

We can find that by saying:

#x^2-2x-15=0#

and now solving for #x#:

#(x-5)(x+3)=0#

#x=5, -3#

And so these are the two values that aren't allowed.

We can see that in the graph:

graph{(x^2-5x+4)/(x^2-2x-15) [-33.28, 39.8, -13.83, 22.75]}

(If you scroll on the graph, you can zoom in on those two #x# values to see they are disallowed).