Question 149237
A)

<b> Domain: </b>


{{{f(x)=(2x^2-18)/(x^2-25)}}} Start with the given function



{{{x^2-25=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-5)(x+5)=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+5=0}}}


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



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




Since {{{x=-5}}} and {{{x=5}}} make the denominator equal to zero, this means we must exclude {{{x=-5}}} and {{{x=5}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq-5 and x\neq5\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>-5}}} or {{{x<>5}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -5\right)\cup\left(-5,5 \right)\cup\left(5,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> -5 and 5 from the domain



--------------------------------------------------



B)

<b> Vertical Asymptote: </b>

To find the vertical asymptote, just set the denominator equal to zero and solve for x


{{{x^2-25=0}}} Set the denominator equal to zero



{{{x^2=0+25}}}Add 25 to both sides



{{{x^2=25}}} Combine like terms on the right side



{{{x=0+-sqrt(25)}}} Take the square root of both sides



{{{x=5}}} or {{{x=-5}}} Simplify



So the vertical asymptotes are {{{x=5}}} or {{{x=-5}}}


--------------------------------------------------


Looking at the numerator {{{2x^2-18}}}, we can see that the degree is {{{2}}} since the highest exponent of the numerator is {{{2}}}. For the denominator {{{x^2-25}}}, we can see that the degree is {{{2}}} since the highest exponent of the denominator is {{{2}}}.



C)

<b> Horizontal/Oblique Asymptote: </b>

Since the degree of the numerator and the denominator are the same, we can find the horizontal asymptote using this procedure:


To find the horizontal asymptote, first we need to find the leading coefficients of the numerator and the denominator.


Looking at the numerator {{{2x^2-18}}}, the leading coefficient is {{{2}}}


Looking at the denominator {{{x^2-25}}}, the leading coefficient is {{{1}}}


So the horizontal asymptote is the ratio of the leading coefficients. In other words, simply divide {{{2}}} by {{{1}}} to get {{{(2)/(1)=2}}}



So the horizontal asymptote is {{{y=2}}}




--------------------------------------------------


D)

<b> Y-Intercept: </b>


To find the y-intercept, simply plug in {{{x=0}}}



{{{y=(2x^2-18)/(x^2-25))}}} Start with the given function



{{{y=(2(0)^2-18)/(0^2-25))}}} Plug in {{{x=0}}}



{{{y=18/25}}} Simplify



So the y-intercept is *[Tex \LARGE \left(0,\frac{18}{25}\right)]



--------------------------------------------------


E)

<b> X-Intercept(s): </b>



To find the x-intercept(s), simply plug in {{{y=0}}} and solve for x



{{{y=(2x^2-18)/(x^2-25))}}} Start with the given function



{{{0=(2x^2-18)/(x^2-25))}}} Plug in {{{y=0}}}



Since the denominator cannot be equal to zero, this means that the numerator is equal to zero.


{{{2x^2-18=0}}}



{{{2x^2=18}}} Add 18 to both sides.



{{{x^2=9}}} Divide both sides by 2.



{{{x=3}}} or {{{x=-3}}} Take the square root of both sides.



So the x-intercepts are  *[Tex \LARGE \left(-3,0\right)] and *[Tex \LARGE \left(3,0\right)]




------------------------------



Notice if we graph {{{y=(2x^2-18)/(x^2-25)}}}, we can visually verify our answers:


{{{drawing(500,500,-10,10,-10,10,
graph(500,500,-10,10,-10,10,(2x^2-18)/(x^2-25)),
blue(line(-20,2,20,2)),
green(line(5,-20,5,20)),
green(line(-5,-20,-5,20))
)}}} Graph of {{{y=(2x^2-18)/(x^2-25))}}}  with the horizontal asymptote {{{y=2}}} (blue line)  and the vertical asymptotes {{{x=-5}}} and {{{x=5}}} (green lines)