Question

find the asymptotes and intercepts of the function. h(x) = 2 / x^3 - 3

Answer 1

$\bf h(x)=\cfrac{2}{x^3-3}\implies h(x)=\cfrac{2x^0}{x^3-3}$

so, the degree of the denominator is higher, then the only horizontal asymptote is y = 0, or the x-axis.

again, to find the vertical ones, zeroing out the denominator.

$\bf x^3-3=0\implies x^3=3\implies x=\sqrt[3]{3}$

and that's the only vertical asymptote.

there are no oblique asymptotes.

again, to get the x-intercept, set y = 0.

$\bf \stackrel{0}{h(x)}=\cfrac{2}{x^3-3}\implies 0=\cfrac{2}{x^3-3}\implies \stackrel{\textit{inconsistent system}}{0\ne 2}$

so, no x-intercepts

let's check for y-intercepts by setting x = 0.

$\bf h(x)=\cfrac{2}{0^3-3}\implies h(x)=-\cfrac{2}{3}$