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
To find the horizontal asymptotes for the given function, we need to examine the behavior of the function as x approaches positive and negative infinity.
First, let's simplify the expression within the square root:
sqr(4x^2 + 3x + 2)
This expression is not equivalent to |x|. Let's correct that:
sqr(4x^2 + 3x + 2) is equivalent to sqr(|x|) or |x|
Now, let's check the limits as x approaches positive and negative infinity:
lim x->+infinity (x-9) / sqr(4x^2 + 3x + 2)
As x goes to positive infinity, the numerator (x-9) also goes to positive infinity. Meanwhile, the denominator (|x|) goes to positive infinity as well.
Therefore, as x approaches positive infinity, the function goes to positive infinity.
lim x->-infinity (x-9) / sqr(4x^2 + 3x + 2)
As x goes to negative infinity, the numerator (x-9) goes to negative infinity, while the denominator (|x|) goes to positive infinity.
Therefore, as x approaches negative infinity, the function goes to negative infinity.
In conclusion, the function does not approach a specific limit from above or below, as the function goes to positive infinity as x approaches positive infinity and goes to negative infinity as x approaches negative infinity.