How do you graph y=(x^2+4x)/(2x-1) using asymptotes, intercepts, end behavior?

1 Answer
Binayaka C.
Jul 18, 2018

Vertical asymptote: x=0.5 , slant asymptote: y= 0.5 x +2.25, y-intercept: (0,0), x intercepts: (0,0), (-4,0), End behavior: x -> -oo , y-> -oo),x -> oo , y-> oo,

Explanation:

y= (x^2+4 x) /(2 x-1)

Vertical asymptote : 2x-1=0 :. x = 1/2 or x =0.5

x -> 0.5^- , y-> - oo and x -> 0.5^+ , y-> oo

Degrees of numerator and denominator are 2 and 1

Since degree of numerator is greater there is slant asymptote

which can be found by long division.

y= 0.5 x + 2.25+ (2.25/(2x-1)) :. Slant asymptote is

y= 0.5 x +2.25 . y intercept: putting x=0 we get

y= (0+0) /(0-1)=0 :. y intercept is at (0,0)

x intercept: putting y=0 we get

:. 0= (x^2+4 x) /(2 x-1) or x(x+4)=0 :. x=0 , x= -4

x intercept is at (0,0), (-4,0)

End behavior : degree is odd, leading co efficient is positive.

For odd degree and positive leading coefficient the graph goes

down as we go left in 3 rd quadrant and goes up as we go

right in 1 st quadrant

Down ( As x -> -oo , y-> -oo),

Up ( As x -> oo , y-> oo),

graph{(x^2+4x)/(2 x-1) [-20, 20, -10, 10]} [Ans]

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