Question 767017
Find the VERT. Asym. Horizontal asymmetry, x-int, y-int for :r(x)= 2+x/16-x^2
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r(x)= 2+x/16-x^2
Horizontal asymptote: y=0 or x-axis
rule: degree of numerator less than degree of denominator.
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Vertical asymptotes:
set denominator=0, then solve for x
16-x^2=0
(4+x)(4-x)=0
x=-4 and x=4
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y-intercept:
set x=0
then solve for y
y=2/16=1/8
y-int=1/8
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x-intercept
set y=0
2+x=0
x=-2

see graph below as a visual check on the solution

{{{ graph( 300, 300, -10, 10, -10, 10,(2+x)/(16-x^2)) }}}