Problem 56 - Algebra and Trigonometry

Question

graph each relation. Use the relation’s graph to determine its domain and range. $$ \frac{y^{2}}{4}-\frac{x^{2}}{25}=1 $$

Step-by-Step Solution

Answer

The domain of the given relation is \((- \infty, \infty)\), and its range is \((- \infty, -2) \cup (2, \infty)\).

Step 1: Graphing the Relation

Start by plotting the hyperbola from the given equation. Because \(a^2 = 4\), the vertical distance from the center to each vertex is \(a = ±2\). And, because \(b^2 = 25\), the horizontal distance from the center to each side of the center rectangle is \(b = ±5\). This leads to a hyperbola opening upwards and downwards.

Step 2: Determine the Domain

The domain of the function is established by the x-values. For a hyperbola of this type, the x-values will span all real numbers because there is no limitation on how wide the hyperbola can go. So, the domain is \( (-\infty, \infty)\).

Step 3: Determine the Range

The range of the function is established by the y-values. Looking at the graph or from the nature of the hyperbola, the y-values present are above \(y = 2\) and below \(y = -2\). Therefore, the range is \((- \infty, -2) \cup (2, \infty)\).

Key Concepts

Domain

When we discuss the domain of a relation, we are referring to all possible x-values that can be input into the equation without breaking any mathematical rules. For a hyperbola like the one given in the exercise: \[ \frac{y^{2}}{4} - \frac{x^{2}}{25} = 1 \]there is no restriction on the x-values. Hyperbolas typically extend indefinitely, allowing x-values to include all real numbers. Thus, for this problem, the domain is all real numbers: \[ (-\infty, \infty). \] This means you can plug in any x-value, and you will still have a valid point on the hyperbola.
Understanding the domain helps ensure that when graphing or calculating points, no errors or undefined situations occur due to invalid x-values.

Range

The range of a relation refers to all possible y-values that can show up when you calculate points on the graph. For this hyperbola, the calculation comes not from direct restriction on y, but rather from the way the hyperbola is shaped.
By analyzing our hyperbola equation: \[ \frac{y^{2}}{4} - \frac{x^{2}}{25} = 1, \]we see the hyperbola opens upward and downward along the y-axis. As a result, for y-values, there's a natural gap in the center of the plot: points above \( y = 2 \)or below\( y = -2 \)exist on the graph, but not between these values. Therefore, the range is written as:\[ (-\infty, -2) \cup (2, \infty). \]
Recognizing these intervals is crucial as they highlight where the hyperbola resides on the y-axis.

Graphing Relations

Graphing a hyperbola involves several steps, starting from understanding its equation and identifying key features. For the equation provided:\[ \frac{y^{2}}{4} - \frac{x^{2}}{25} = 1, \]this is a standard form of a hyperbola that opens vertically. Recognizing this form helps decide how the hyperbola is oriented.
From here, we identify the hyperbola's center, vertices, and the axes distances; specifically, values of \( a \)and \( b \).Here, \( a = 2 \)(taken from\( a^2 = 4 \))maps out the vertical reach from the center, while \( b = 5 \)(derived from \( b^2 = 25 \))shows the horizontal extent.
When sketching, the vertices will sit above and below the center on the y-axis, at points \( (0, ±2) \).Aided with these key points and dimensions, plot the hyperbola always ensuring it shapes as two "mirror-image" curves.
This visual outline permits the finding of both domain and range, emphasizing the graph's role in understanding the relation completely.