Question

determine the vertical asymptotes of the graph of the function. give your answers as equations in exact form.

$$ k ( t ) = \frac { t ^ { 4 } - 1 } { 3 t ^ { 2 } + 8 t - 1 } $$

separate multiple equations with commas as necessary. select
one\ if applicable.

equation(s) of the vertical asymptote(s):

Explanation:

Step1: Find the denominator of the rational function

The denominator of the function \(k(t)=\frac{t^{4}-1}{3t^{2}+8t - 1}\) is \(y = 3t^{2}+8t - 1\).

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c=0\) (\(a = 3\), \(b = 8\), \(c=-1\)), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
Substitute the values: \(t=\frac{-8\pm\sqrt{8^{2}-4\times3\times(-1)}}{2\times3}=\frac{-8\pm\sqrt{64 + 12}}{6}=\frac{-8\pm\sqrt{76}}{6}=\frac{-8\pm2\sqrt{19}}{6}=\frac{-4\pm\sqrt{19}}{3}\).

Step3: Write the equations of vertical asymptotes

The vertical asymptotes of a rational function \(y=\frac{f(t)}{g(t)}\) occur at the values of \(t\) that make \(g(t)=0\) (provided the numerator \(f(t)
eq0\) at those values). Since \(t^{4}-1
eq0\) when \(3t^{2}+8t - 1 = 0\), the equations of the vertical asymptotes are \(t=\frac{-4+\sqrt{19}}{3}\) and \(t=\frac{-4-\sqrt{19}}{3}\).

Answer:

$t=\frac{-4+\sqrt{19}}{3},t=\frac{-4-\sqrt{19}}{3}$