Question

graph the hyperbola described by the equation: \\(\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{4} = 1\\) graph grid and drawing tools as shown question help: message instructor next question

Explanation:

Step1: Identify Hyperbola Type

The standard form of a vertical hyperbola is \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\), where \((h,k)\) is the center. Here, the equation is \(\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{4} = 1\), so it's a vertical hyperbola (since the \(y\)-term is positive). Center \((h,k)=(-1,1)\) (from \(x+1 = x - (-1)\) and \(y - 1\)).

Step2: Find \(a\), \(b\), and Asymptotes

For \(\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{4} = 1\), \(a^2 = 16\) so \(a = 4\), \(b^2 = 4\) so \(b = 2\). The asymptotes for a vertical hyperbola are \(y - k=\pm\frac{a}{b}(x - h)\), so \(y - 1=\pm\frac{4}{2}(x + 1)\) or \(y - 1=\pm2(x + 1)\).

Step3: Plot Center and Vertices

Center is \((-1,1)\). Vertices are \(a\) units above and below the center: \((-1,1 + 4)=(-1,5)\) and \((-1,1 - 4)=(-1,-3)\).

Step4: Draw the Hyperbola

Using the center, vertices, and asymptotes (the dashed lines in the graph match \(y - 1=\pm2(x + 1)\)), the hyperbola opens up and down (vertical), so we plot the two branches above and below the center, using the asymptotes as guides. The left - most draw option (the one with the vertical opening, like two curves opening up and down) should be used.

Answer:

To graph the hyperbola \(\frac{(y - 1)^2}{16} - \frac{(x + 1)^2}{4}=1\):

1. Center: \((-1,1)\) (from \(x+1=x - (-1)\) and \(y - 1\)).
2. Type: Vertical hyperbola (since the \(y\)-term is positive in the standard form \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\)).
3. Vertices: \(a = 4\) units from the center along the \(y\)-axis: \((-1,1 + 4)=(-1,5)\) and \((-1,1 - 4)=(-1,-3)\).
4. Asymptotes: \(y - 1=\pm2(x + 1)\) (matches the dashed lines in the graph).
5. Graph: Use the "vertical hyperbola" draw option (the left - most of the two draw icons, with branches opening up and down) to plot the two branches centered at \((-1,1)\), passing through the vertices, and guided by the asymptotes.

(Note: The final action is to draw the hyperbola with the correct shape (vertical, center \((-1,1)\), vertices \((-1,5)\) and \((-1,-3)\), and asymptotes \(y - 1=\pm2(x + 1)\)) using the appropriate draw tool.)