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]

Related questions
See all questions in Limits - End Behavior and Asymptotes
Impact of this question
2122 views around the world
You can reuse this answer
Creative Commons License