determine the vertical asymptotes of the graph of the function.\ng(x)=\\frac{x^{3}}{2 x^{3}-x^{2}-36…
determine the vertical asymptotes of the graph of the function.\ng(x)=\\frac{x^{3}}{2 x^{3}-x^{2}-36 x}\nselect the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.\na. the function has one vertical asymptote, \n(type an equation. use integers or fractions for any numbers in the equation.)\nb. the leftmost vertical asymptote is and the rightmost vertical asymptote is \n(type equations. use integers or fractions for any numbers in the equations.)\nc. the function has no vertical asymptotes.
Answer
Explanation:
Step1: Factor numerator and denominator
Factor numerator: $x^3=x\cdot x\cdot x$. Factor denominator: $2x^3 - x^2 - 36x=x(2x^2 - x - 36)$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ for $ax^2+bx + c = 0$ (here $a = 2$, $b=-1$, $c=-36$), we have $x=\frac{1\pm\sqrt{1-4\times2\times(-36)}}{2\times2}=\frac{1\pm\sqrt{1 + 288}}{4}=\frac{1\pm\sqrt{289}}{4}=\frac{1\pm17}{4}$. So $2x^2 - x - 36=2(x-\frac{1 + 17}{4})(x-\frac{1-17}{4})=2(x-\frac{9}{2})(x + 4)$. The function becomes $g(x)=\frac{x^3}{x(2x^2 - x - 36)}=\frac{x^2}{2(x-\frac{9}{2})(x + 4)}$ ($x\neq0$).
Step2: Find vertical asymptotes
Vertical asymptotes occur where the denominator is zero (after canceling common factors). Set $2(x-\frac{9}{2})(x + 4)=0$, we get $x=\frac{9}{2}$ and $x=-4$.
Answer:
B. The leftmost vertical asymptote is $x=-4$ and the rightmost vertical asymptote is $x=\frac{9}{2}$.