Answer:
Step-by-step explain
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
Answer:
you could explain how to get the answer and add detailed words to make it more like english form
Step-by-step explanation:
ok so first you would find the area and perimeter and state how you got your answer
Answer: the graph farthest to the right is almost correct. If you substitute values for x in the function f(x)= -3√x , the output does not match the curve on the graphs shown.
If you have a choice that includes only a curve to the right of the y- axis, that would be better.
Step-by-step explanation: Square roots of Negative x-values will result in imaginary numbers. Otherwise the graph with the curve passing through coordinates (1,-3) (4,-6) and (9,-9) is a good choice.
(And ask your teacher about the square root of negative numbers on this graph.)
Answer:
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Step-by-step explanation:
r u sure thats college work
