SOLUTION: Given {{{y=(2x^2-18)/(x^2-25)}}}, A) Find the domain. B) Determine the vertical asymptote(s). C) Determine the horizontal asymptote or oblique asymptote. D) Find th

Algebra ->  Graphs -> SOLUTION: Given {{{y=(2x^2-18)/(x^2-25)}}}, A) Find the domain. B) Determine the vertical asymptote(s). C) Determine the horizontal asymptote or oblique asymptote. D) Find th      Log On


   

Question 149237: Given y=%282x%5E2-18%29%2F%28x%5E2-25%29,

A) Find the domain.
B) Determine the vertical asymptote(s).
C) Determine the horizontal asymptote or oblique asymptote.
D) Find the y-intercept.
E) Find the x-intercept(s).
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A)
Domain:

f%28x%29=%282x%5E2-18%29%2F%28x%5E2-25%29 Start with the given function


x%5E2-25=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.




%28x-5%29%28x%2B5%29=0 Factor the left side (note: if you need help with factoring, check out this solver)




Now set each factor equal to zero:

x-5=0 or x%2B5=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:

which in plain English reads: x is the set of all real numbers except x%3C%3E-5 or x%3C%3E5

So our domain looks like this in interval notation


note: remember, the parenthesis excludes -5 and 5 from the domain


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B)
Vertical Asymptote:
To find the vertical asymptote, just set the denominator equal to zero and solve for x

x%5E2-25=0 Set the denominator equal to zero


x%5E2=0%2B25Add 25 to both sides


x%5E2=25 Combine like terms on the right side


x=0%2B-sqrt%2825%29 Take the square root of both sides


x=5 or x=-5 Simplify


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

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Looking at the numerator 2x%5E2-18, we can see that the degree is 2 since the highest exponent of the numerator is 2. For the denominator x%5E2-25, we can see that the degree is 2 since the highest exponent of the denominator is 2.


C)
Horizontal/Oblique Asymptote:
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%5E2-18, the leading coefficient is 2

Looking at the denominator x%5E2-25, 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 %282%29%2F%281%29=2


So the horizontal asymptote is y=2



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D)
Y-Intercept:

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


y=%282x%5E2-18%29%2F%28x%5E2-25%29%29 Start with the given function


y=%282%280%29%5E2-18%29%2F%280%5E2-25%29%29 Plug in x=0


y=18%2F25 Simplify


So the y-intercept is


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E)
X-Intercept(s):


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


y=%282x%5E2-18%29%2F%28x%5E2-25%29%29 Start with the given function


0=%282x%5E2-18%29%2F%28x%5E2-25%29%29 Plug in y=0


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

2x%5E2-18=0


2x%5E2=18 Add 18 to both sides.


x%5E2=9 Divide both sides by 2.


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


So the x-intercepts are and



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Notice if we graph y=%282x%5E2-18%29%2F%28x%5E2-25%29, we can visually verify our answers:

Graph of y=%282x%5E2-18%29%2F%28x%5E2-25%29%29 with the horizontal asymptote y=2 (blue line) and the vertical asymptotes x=-5 and x=5 (green lines)