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State the horizontal asymptote of the rational function.

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numerator degree is 1, denominator has degree 2 and since 2 is larger than 1 the horizontal asymptote is \(y=0\)
for the vertical asymptotes, set the denominator equal to zero and solve for \(x\)
i dont get how you got y=o pplz help with horizontal

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if the degree of the numerator is smaller than the degree of the denominator, you will have a horizontal asymptote at \(y=0\) aka the \(x\)- axis do you know what i mean by degree?
yeah i know what degree is
what if the degree of the numnater is bigger
ok then that is all you need to consider think about what would happen if you put in say \(x=10^6\) you would have a million in the numerator but you would have \(10^{12}\) in the denominator, and get a very small number (close to zero)
if the degree of the numerator is larger, no horizontal asymptote
and if the degrees are the same, it is the ratio of the leading coefficients
so its allways y=0 when the top is smaller and none when the top is larger
yes, always
and if they are the same, for example \[\frac{2x^2+3x-3}{5x^2+10}\] i would be \(y=\frac{2}{5}\)
why 5?
you got that? three cases 1) top is larger: none 2) bottom is larger :\(y=0\) 3) the degrees are equal: \(y=\text{ratio of leading coefficients}\)
in my example \[\frac{2x^2+3x-3}{5x^2+10}\] the degrees are both two the leading coefficient of the numerator is 2 the leading coefficent of the denominator is 5 so horizontal asymptote is \(y=\frac{2}{5}\)
State the horizontal asymptote of the rational function. 9x^2-3x-8 --------- 4x^2-5x+3 so this would be y=9/4?
sure would
thanks your the best!! can you fan me so i can message you if i need help plz?
you have to admit this is rather easy, right? i mean once you know what you are doing
yeah i posted this because i am home schooled and i have no one to teach me the concepts
that must be tough guess you can always post here. there are lots of on line resources, but here you may get a direct answer

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