Question

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

Answer 1

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]