Answer:
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Find the horizontal asymptote for f(x)=(3x^2-1)/(2x-1) :
A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of y=a_n/b_n if the degree of the numerator is the same as the degree of the denominator (where a_n,b_n are the leading coefficients of the numerator and denominator respectively when both are in standard form.)
If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.
For the given function, there is no horizontal asymptote.
We can find the slant asymptote by using long division:
(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))
The slant asymptote is y=3/2x+3/4
Your answer is 110. You take 10 and divide it by 750. = 75 then you take 75. Divided 8250. = 110
Answer:
huh?
Step-by-step explanation:
