please help with asymptotes....

Question

lovelyharmonics · Answer

f(x)=(7x^2-3x-9)/(2x^2-4x+5)

lovelyharmonics · Answer

State the horizontal asymptote of the rational function.

smrougraff · Answer

try putting it in a graphing calculator and see which horizontal lines it doesn't go threw or factor it out and cancel out then the ones left can get you your answer

lovelyharmonics · Answer

@ganeshie8 @satellite73 help :c my graph looks like a square root sign decided not to stop and just kept going :(

lovelyharmonics · Answer

the first ones is whats showing up on my calculator.....

lovelyharmonics · Answer

with what? 7,3,and9 have nothing to do with eachother
same goes for 2,4,and5 .-.
like 3 is a factor of 9 but you cant just stick 7 out on the end

anonymous · Answer

hmmm
Remove everything except the terms with the biggest exponents of x found in the numerator and denominator

anonymous · Answer

so you get f(x) = 7x^2/2x^2 =7/2

lovelyharmonics · Answer

okay.... and what does that do .-.

anonymous · Answer

so 7/2 will be ur horizontal asymptote .-.

lovelyharmonics · Answer

but 7/2 would imply that my horizontal asymptote would be 3 1/2 but my graph goes above that .-.

anonymous · Answer

http://www.wolframalpha.com/input/?i=f%28x%29%3D%287x%5E2-3x-9%29%2F%282x%5E2-4x%2B5%29+horizontal+asymptote

anonymous · Answer

see the result

lovelyharmonics · Answer

how ddoes that work D: the graph isnt suppose to pass the asymptote

anonymous · Answer

no
when x is infinity,then they are asking y's value

anonymous · Answer

it doesnt matter what value is it crossing
what matters is,what value is it giving when x = infinity

anonymous · Answer

so when x is infinity
then y's value is near 7/2
thats why its the horizontal asymtote

anonymous · Answer

got it?

anonymous · Answer

its not that hard to get .-. @lovelyharmonics

lovelyharmonics · Answer

yeah definitely c: thank you

anonymous · Answer

welcome