the graph shows a hyperbola. what is its eccentricity? write your answer in simplified, rationalized form.

Explanation:

Step1: Identify the standard - form of hyperbola

The hyperbola is vertical with center at the origin $(0,0)$. The general equation of a vertical hyperbola centered at the origin is $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1$. The vertices are at $(0,\pm a)$. From the graph, the vertices are at $(0,6)$ and $(0, - 6)$, so $a = 6$.

Step2: Find the slope of the asymptotes

The equations of the asymptotes of a vertical hyperbola $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ are $y=\pm\frac{a}{b}x$. The slope of the asymptotes from the graph is $\frac{a}{b}=3$ (by observing the rise - over - run of the asymptotes). Since $a = 6$ and $\frac{a}{b}=3$, we can solve for $b$: $6 = 3b$, so $b = 2$.

Step3: Use the eccentricity formula

The eccentricity $e$ of a hyperbola is given by the formula $e=\sqrt{1+\frac{b^{2}}{a^{2}}}$. Substitute $a = 6$ and $b = 2$ into the formula: $e=\sqrt{1+\frac{2^{2}}{6^{2}}}=\sqrt{1+\frac{4}{36}}=\sqrt{\frac{36 + 4}{36}}=\sqrt{\frac{40}{36}}=\frac{\sqrt{40}}{6}=\frac{2\sqrt{10}}{6}=\frac{\sqrt{10}}{3}$.

Answer:

$\frac{\sqrt{10}}{3}$