Question

match the equation with the graph. $\frac{y^{2}}{16}-\frac{x^{2}}{1}=1$

Explanation:

Step1: Recall the standard form of a hyperbola

The standard form of a hyperbola is \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\) (opens up - down) or \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) (opens left - right). For the equation \(\frac{y^{2}}{1}-\frac{x^{2}}{16}=1\), it is of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}} = 1\) where \(a = 1\) and \(b = 4\). This hyperbola opens up - down.

Step2: Analyze the vertices

For a hyperbola \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\), the vertices are at \((0,\pm a)\). Here \(a = 1\), so vertices are \((0,1)\) and \((0, - 1)\). The first graph has vertices near \((0,\pm1)\) and is a hyperbola of the form \(\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1\). The second graph is a hyperbola of the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) (opens left - right). The third graph is a pair of parabolas. The fourth graph is a combination of a parabola opening up and a parabola opening down.

Answer:

The first graph (the one with vertices near \((0,\pm1)\) and opening up - down) is the graph of the equation \(\frac{y^{2}}{1}-\frac{x^{2}}{16}=1\)