Asked by brian
find the HAs of
y = (x-9) / sqr(4x^2 + 3x+ 2)
What happens to y as x goes to +/- infinity? Specifically, is y approaching some limit from above or below?
??
I see your question was to just find the horizontal asymptotes and not to evaluate the function's approach to them.
The denominator has no real zeroes, so it's always positive. For large x the denominator behaves like |x| and the numerator is just x. The limits that need to be checked are
lim x-> +/-infinity x/|x|
Does that help?
so
sqr(4x^2 + 3x +2) is equivalent to |x|
Check that, it for large x it behaves like 2|x| -I missed the 4.
thx
y = (x-9) / sqr(4x^2 + 3x+ 2)
What happens to y as x goes to +/- infinity? Specifically, is y approaching some limit from above or below?
??
I see your question was to just find the horizontal asymptotes and not to evaluate the function's approach to them.
The denominator has no real zeroes, so it's always positive. For large x the denominator behaves like |x| and the numerator is just x. The limits that need to be checked are
lim x-> +/-infinity x/|x|
Does that help?
so
sqr(4x^2 + 3x +2) is equivalent to |x|
Check that, it for large x it behaves like 2|x| -I missed the 4.
thx
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