Asked by Kelsi
                hi! The question is: a.)identify any horizontal and vertical asymptote and (b) identidify any holes in the graph. Verify your answers numerically by creating a table of values. 
14.) f(x)= 3/(x-2)^3
16.) f(x)= x^2+2x+1)/2x^2-x-3
18.) 3-14x-5x^2)/3+7x+2x^2
I'm not looking for the answer necessarily, but I have no idea how to do this and the steps in my book aren't helping. Is there a way that whoever helps me can walk me through each step for each problem? Pretend you are teaching this to a 4th grader! Thank you!
            
        14.) f(x)= 3/(x-2)^3
16.) f(x)= x^2+2x+1)/2x^2-x-3
18.) 3-14x-5x^2)/3+7x+2x^2
I'm not looking for the answer necessarily, but I have no idea how to do this and the steps in my book aren't helping. Is there a way that whoever helps me can walk me through each step for each problem? Pretend you are teaching this to a 4th grader! Thank you!
Answers
                    Answered by
            Damon
            
    if a denominator is zero, you have a vertical asymptote (look at 14 as x becomes 2)
if the function becomes constant you have a horizontal. look at 14 as x gets big positive and as x gets big negative
    
if the function becomes constant you have a horizontal. look at 14 as x gets big positive and as x gets big negative
                    Answered by
            Kelsi
            
    Ok that helps with the asymptotes... Could you help me with everything else? 
    
                    Answered by
            Steve
            
    holes occur where the numerator and denominator are both zero.
#16
x^2+2x+1 = (x+1)(x+1)
2x^2-x-3 = (x+1)(2x-3)
So, f(x) = (x+1)/(2x-3)
everywhere except at x = -1. At that point, f(x) = 0/0 which is not defined. The only vertical asymptote is at x = 3/2, and at x = -1 there is a hole. No matter how close you get to x = -1, f(x) gets very very close to 0, but at exactly x = -1, f(x) is undefined.
As x gets huge, f(x) is very close to x^2/2x^2 = 1/2, so y = 1/2 is the horizontal asymptote.
See the graph at
http://www.wolframalpha.com/input/?i=%28x^2%2B2x%2B1%29%2F%282x^2-x-3%29
    
#16
x^2+2x+1 = (x+1)(x+1)
2x^2-x-3 = (x+1)(2x-3)
So, f(x) = (x+1)/(2x-3)
everywhere except at x = -1. At that point, f(x) = 0/0 which is not defined. The only vertical asymptote is at x = 3/2, and at x = -1 there is a hole. No matter how close you get to x = -1, f(x) gets very very close to 0, but at exactly x = -1, f(x) is undefined.
As x gets huge, f(x) is very close to x^2/2x^2 = 1/2, so y = 1/2 is the horizontal asymptote.
See the graph at
http://www.wolframalpha.com/input/?i=%28x^2%2B2x%2B1%29%2F%282x^2-x-3%29
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