Question 200397: i need to find the domain, intercepts and any asymptotes of f(x)=(x^2+3X-4)/(x+1) but when i graph it it's a hyperbola and its throwing me off!! please help!!
Answer by RAY100(1637)   (Show Source):
You can put this solution on YOUR website! y= (x^2 +3x -4 )/ (x+1)
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y= { ( x+4)(x-1) } / (x+1)
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x intercepts, zeros of num,,,x=-4,,,x=+1
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y intercepts , at x=0,,,,,,,y = -4
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Vertical Asymptote,,, zero of den ,,,,,x=-1
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Horizontal Asymptote, none,, x^2 > x ( Num & den)
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However there is a slant asymptote,,,( x^2 one power higher than x)
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,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,x,,,,+2,,,,,,,,,,,,,,,r=-6
,,,,,,,,,,,,,,,,,,,,,,--------------------------
(x+1),,,,,,,,,,,,,!,,,x^2 +3x,,,,,-4
,,,,,,,,,,,,,,,,,,,,,,,,,x^2 +x
,,,,,,,,,,,,,,,,,,,,,,,,,_________
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2x -4
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2x +2
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,_______
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,-6
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Slant asymptote is,,,,,,,y= x+2
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To get a rough sketch , use x-y coordinates, lay in intercepts, y=-4,x=-4,x=+1
Show vertical asymptote at x=-1, and slant asymptote along y =x+2
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Starting at left, curve is just above slant asy,, rises thru x=-4, and continues up at vert asy at x=-1. On the other side of vert asy, it is at = - infinity, rising thru (0,-4) and (1,0)
to slant asy.
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