Range of 5/(x-3)

its got a va at 3 so id assume the range is all up and down

except for 0 maybe

The horizontal asymptote is y = 0, so this is the only output value that y cannot equal is 0 So the range is the set of all real numbers but y cannot equal 0

kinda hard to equal 0 when you got a constant up top :)

no, if the degree of the denominator exceeds the degree of the numerator, then the horizontal asymptote is always y = 0

I'm not doing equating of any kind, I'm finding the horizontal asysmptote

well yeah, if you wanna go by rules and not logic...

.... that was peculiar statement; since the rules are founded on logic ;)

5/x = 0 when 5 = 0, so never

did asymptotes last year and cannot remember them. how do you fing the horizantal asymptote

logically, this is explained by the fact that the degree has a higher rate of growth, so as x approaches infinity, the denominator will be much much larger than the numerator So this means we'll have a very small fraction (when x --> oo)

ideally, calculus is needed to see this, but you can see this visually too

3 simple rules for HA\[\] bottom bigger than top; 0\[\] bottom = top; leading coeffs\[\] bottom smaller than top; divide out and toss the remainder

oops meant to write "denominator has higher rate of growth"

exactly, so what's the issue?

march 1984 SI edition of sports illustrated

swimsuit edition that is

just to butt in at the last minute, not sure calc is needed. it is pretty clear that \[\frac{5}{x-3}\] is never 0 since a fraction is only 0 if the numerator is

it approaches zero though as x approaches infinity

k... thanks

now now, i think we might need calculus to determine this for us; lets see what leibniz and newton are up to these days :)

they'll fight to the death like they did 300 yrs ago...so idk if they'll be of much help lol

lol ... :)

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