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

Jul 18, 2018

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

#### Explanation:

$y = \frac{{x}^{2} + 4 x}{2 x - 1}$

Vertical asymptote : $2 x - 1 = 0 \therefore x = \frac{1}{2} \mathmr{and} x = 0.5$

$x \to {0.5}^{-} , y \to - \infty \mathmr{and} x \to {0.5}^{+} , y \to \infty$

Degrees of numerator and denominator are $2 \mathmr{and} 1$

Since degree of numerator is greater there is slant asymptote

which can be found by long division.

$y = 0.5 x + 2.25 + \left(\frac{2.25}{2 x - 1}\right) \therefore$ Slant asymptote is

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

$y = \frac{0 + 0}{0 - 1} = 0 \therefore$ y intercept is at $\left(0 , 0\right)$

x intercept: putting $y = 0$ we get

$\therefore 0 = \frac{{x}^{2} + 4 x}{2 x - 1} \mathmr{and} x \left(x + 4\right) = 0 \therefore x = 0 , x = - 4$

x intercept is at $\left(0 , 0\right) , \left(- 4 , 0\right)$

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 $x \to - \infty , y \to - \infty$),

Up ( As $x \to \infty , y \to \infty$),

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