Question 110923
{{{f(x)=(x^2-3x-10)/(x^2-6x+9)}}} Start with the given function



{{{x^2-6x+9=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.





{{{(x-3)(x-3)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)





Now set each factor equal to zero:


{{{x-3=0}}} or {{{x-3=0}}}


{{{x=3}}} or {{{x=3}}}  Now solve for x in each case



Since these two answers are the same, our only solution is {{{x=3}}}




Since {{{x=3}}} makes the denominator equal to zero, this means we must exclude {{{x=3}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq3 \right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>3}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, 3\right)\cup\left(3,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> 3 from the domain




To find the x-intercepts, set the entire expression equal to zero


{{{0=(x^2-3x-10)/(x^2-6x+9)}}}



So this means the numerator can only equal zero


{{{x^2-3x-10=0}}}




{{{(x-5)(x+2)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:


{{{x-5=0}}} or {{{x+2=0}}}


{{{x=5}}} or {{{x=-2}}}  Now solve for x in each case



So our solutions are {{{x=5}}} or {{{x=-2}}}


This means the x intercepts are (-2,0) or (5,0)



Notice if we graph {{{f(x)=(x^2-3x-10)/(x^2-6x+9)}}}, we can see that the x intercepts are (-2,0) or (5,0) and {{{x=3}}} is not in the domain
 


{{{ graph( 500, 500, -10, 10, -10, 10, (x^2-3x-10)/(x^2-6x+9)) }}}



So this verifies our answer.