Question 449350: 1)Given the function y=x+2/2x^2-4
what are the
a)x intercept
b)y intercept
c)vertical asymptote(s)
d)horizontal asymptote(s)
e)holes
f)graph of how it looks
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 1)Given the function y=x+2/2x^2-4
what are the
a)x intercept
b)y intercept
c)vertical asymptote(s)
d)horizontal asymptote(s)
e)holes
f)graph of how it looks
..
a) x-intercept
To find the x-intercepts, set y=0, then solve for x.
Note that when y=0, equating the numerator=0, will satisfy the equation
For given function,
x+2=0
x=-2 (x-intercept)
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b) y-intercept
To find y-intercepts, set x=0, then solve for y
y=2/-4=-1/2 (y-intercept)
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c)vertical asymptote(s)
To find vertical asymptotes, set denominator=0, then solve for x.
2x^-4=0
2x^2=4
x^2=2
x=±√2
Vertical asymptotes are equations of lines, that is, the two asymptotes are x=√2 and x=-√2
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d)horizontal asymptote(s)
When the degree of the numerator is less than that of the denominator,as in given case, the horizontal asymptote is the x-axis or y=0. If the degrees of numerator and denominator are the same,not in this case, divide the coefficient of the x term in the numerator by the x term in denominator. The quotient of this division becomes the horizontal asymptote.
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e)holes: There are no holes in this rational equation.
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f)graph of how it looks
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