# How do I identify the x-intercept(s) and vertical asymptote(s): y=(x^3+27)/(3x^2+x)?

Feb 26, 2015
1. To identify the x-intercepts, you want to ask yourself: "where does the graph hit the x-axis, aka: what is x when y=0?"
So let y = 0 and solve for x:
$0 = \frac{{x}^{3} + 27}{3 {x}^{2} + x}$
In order for this fraction to equal 0, the numerator of the fraction must equal 0 (remember: denominator = 0 -> undefined)
$0 = \left({x}^{3} + 27\right)$
${x}^{3} = - 27$
$x = - 3$
So the x-intercept: (-3,0)

2. To identify the vertical asymptotes, we first try and simplify the function as much as possible and then look at where it is undefined
$y = \frac{{x}^{3} + 27}{3 {x}^{2} + x}$ is already simplified
Undefined when denominator = 0: $\left(3 {x}^{2} + x\right) = 0$
$x \left(3 x + 1\right) = 0$
$x = 0 , 3 x + 1 = 0$
Vertical asymptotes: $x = 0 , x = - \frac{1}{3}$