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
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!
3 answers
Ok that helps with the asymptotes... Could you help me with everything else?
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