Problem 57 - A Graphical Approach to College Algebra

Question

Graph each hyberbola by hand. Give the domain and range. Do not use a calculator. $$\frac{(y-5)^{2}}{4}-\frac{(x+1)^{2}}{9}=1$$

Step-by-Step Solution

Answer

Domain: all real numbers; Range: \((- fty, 3) \cup (7, fty)\).

Step 1: Identify the Hyperbola Type

The given hyperbola equation is \( \frac{(y-5)^{2}}{4} - \frac{(x+1)^{2}}{9} = 1 \). This is in the form \( \frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1 \), which indicates it is a vertical hyperbola.

Step 2: Locate the Center

The center of the hyperbola is found from \((x+1)^2\) and \((y-5)^2\). This results in the center at the point \((-1, 5)\).

Step 3: Calculate the Vertices

For a vertical hyperbola, vertices are located \(a\) units from the center along the y-axis. Since \(a^2 = 4\), \(a = 2\). Thus, the vertices are \((-1, 5 \pm 2)\) or \((-1, 3)\) and \((-1, 7)\).

Step 4: Calculate the Co-Vertices

Co-vertices are located \(b\) units from the center along the x-axis. Here, \(b^2 = 9\), so \(b = 3\). The co-vertices are \((-1 \pm 3, 5)\) or \((-4, 5)\) and \((2, 5)\).

Step 5: Draw the Asymptotes

The asymptotes for a vertical hyperbola centered at \((h, k)\) are \(y = k \pm \frac{a}{b}(x-h)\). Substituting \(a = 2\), \(b = 3\), and the center \((-1, 5)\), the equations are \(y = 5 \pm \frac{2}{3}(x + 1)\).

Step 6: Graph the Hyperbola

Plot the center at \((-1, 5)\), the vertices at \((-1, 3)\) and \((-1, 7)\), and draw the asymptotes using the equations \(y = 5 + \frac{2}{3}(x + 1)\) and \(y = 5 - \frac{2}{3}(x + 1)\). Sketch the hyperbola branches opening upwards and downwards around the asymptotes.

Step 7: Determine the Domain and Range

The domain of a vertical hyperbola is all real numbers, since the value of \(x\) is not restricted. The range is \((-fty, 3) \cup (7, fty)\) because the hyperbola opens upwards and downwards from the vertices at \(y = 3\) and \(y = 7\).

Key Concepts

Vertical Hyperbola

A vertical hyperbola, like the one given in the equation \( \frac{(y-5)^{2}}{4} - \frac{(x+1)^{2}}{9} = 1 \), opens up and down. This is because the variable \(y\) is associated with the positive term of the equation, while \(x\) is associated with the negative. This specific form is known as the standard form for vertical hyperbolas.In a vertical hyperbola, the center, vertices, and co-vertices are essential points that help plot the graph. The graph of a vertical hyperbola is shaped like two opposing curves (often compared to a saddle), with each curve moving away from the center upwards and downwards. The curves approach but never actually intersect the asymptotes, which provide an imagined boundary."},{