Question 723383: state what kind of asymptotes the function has, find their equations and find the zeros.
f(x)=(2x^2-3)/(x+4)
i dont even know where to start for this one
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! state what kind of asymptotes the function has, find their equations and find the zeros.
f(x)=(2x^2-3)/(x+4)
Because the degree of the numerator is one degree higher than that of the denominator, the function has a slant or oblique asymptote. To find it, divide numerator by denominator by long division. The quotient, ignoring the remainder, is the line equation of the slant asymptote.
(2x^2-3)/(x+4)=(x-8)+Remainder=29
..
To find vertical asymptotes, set denominator=0, then solve for x-values that make the function undefined.
x+4=0
x≠-4
..
To find the zeros, set the function=0, which makes the numerator=0
2x^2-3=0
2x^2=3
x^2=3/2
x=±√(3/2)
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