Problem 2
Question
Make a table of values. Then sketch a graph of each inverse variation. \(y=\frac{10}{x}\)
Step-by-Step Solution
Verified Answer
For the inverse function \(y= \frac{10}{x}\), the table of values is: (-3, -10/3), (-2, -5), (-1, -10), (1, 10), (2, 5), (3, 10/3). The graph of the function is a hyperbola located in the first and third quadrants.
1Step 1: Set up a table of values
Start by selecting a range of meaningful values for x to be substituted into the function, ensuring to include both positive and negative values. In this case, use -3, -2, -1, 1, 2, 3 for x. Substitute all these values into the function and obtain the corresponding y values.
2Step 2: Calculate y values
Next, calculate the y values for each x by substituting the x values into the function. So when x=-3, y= \(-\frac{10}{3}\); for x=-2, y= \(-\frac{10}{2}\) or -5; for x=-1, y= \(-\frac{10}{1}\) or -10; for x=1, y=\(\frac{10}{1}\) or 10; for x=2, y= \(\frac{10}{2}\) or 5; and finally when x=3, y= \(\frac{10}{3}\).
3Step 3: Draw the graph
Finally, plot the points on the graph: (-3, -10/3), (-2, -5), (-1, -10), (1, 10), (2, 5), and (3, 10/3). The shape of the graph would be a hyperbola, demonstrating the inverse variation between x and y. There will be two curves, one occurring in the first quadrant and the other in the third quadrant of the graph.
Key Concepts
Graphing FunctionsHyperbolaAlgebra 2
Graphing Functions
Learning how to graph functions is a fundamental skill in Algebra 2, and it helps in understanding the behavior of different mathematical relationships. When graphing functions, it is important to start by determining a set of values for the variable \(x\). By substituting these values into the function, you can compute the corresponding \(y\) values. This process creates a table of values useful for plotting points on a graph.
For the function \(y = \frac{10}{x}\), we choose a range of \(x\) values such as \(-3, -2, -1, 1, 2,\) and \(3\). Substitute each into the equation to find \(y\). This will give points like \((-3, -10/3)\), \((-2, -5)\), and \((1, 10)\). After plotting these points, you'll see how they relate to one another and start forming a shape that is characteristic of an inverse variation graph.
For the function \(y = \frac{10}{x}\), we choose a range of \(x\) values such as \(-3, -2, -1, 1, 2,\) and \(3\). Substitute each into the equation to find \(y\). This will give points like \((-3, -10/3)\), \((-2, -5)\), and \((1, 10)\). After plotting these points, you'll see how they relate to one another and start forming a shape that is characteristic of an inverse variation graph.
- Ensure to pick both negative and positive values for a full graph depiction.
- The result will be two separate curves showing the inverse relationship.
- Always make sure the graph fits the pattern described by the equation.
Hyperbola
A hyperbola isn't the type of curve you might see every day, but it's an important shape in mathematics, especially in the context of inverse variation. Recognizing the shape of a hyperbola helps you understand how different variables interact inversely.
In an inverse variation graph like the one for \(y = \frac{10}{x}\), the resulting graph manifests as a hyperbola. This happens because as \(x\) increases, \(y\) decreases proportionally, and vice versa. The hyperbola will always consist of two opposite-facing curves: one in the first quadrant and another in the third quadrant.
In an inverse variation graph like the one for \(y = \frac{10}{x}\), the resulting graph manifests as a hyperbola. This happens because as \(x\) increases, \(y\) decreases proportionally, and vice versa. The hyperbola will always consist of two opposite-facing curves: one in the first quadrant and another in the third quadrant.
- The curves never touch the axes. They approach them indefinitely, known as asymptotes.
- This graph shows a relationship where the product of \(x\) and \(y\) equals a constant (in this case, 10).
- This specific form of hyperbola is always symmetric around the origin.
Algebra 2
Algebra 2 is a high school mathematics course that builds upon algebraic concepts from earlier coursework. It dives deeper into the analysis of functions, including linear, quadratic, and especially rational functions like inverse variations.
In the context of inverse variation, Algebra 2 introduces students to more complex relationships. The equation \(y = \frac{k}{x}\) is a typical start, where \(k\) is a constant. Understanding how this variation works is crucial because it appears in many applications—science, economics, and even engineering. In Algebra 2:
In the context of inverse variation, Algebra 2 introduces students to more complex relationships. The equation \(y = \frac{k}{x}\) is a typical start, where \(k\) is a constant. Understanding how this variation works is crucial because it appears in many applications—science, economics, and even engineering. In Algebra 2:
- You learn how changes in one variable impact another inversely.
- Identifying parameters like asymptotes and intercepts becomes second nature.
- Graphing these relationships helps in visualizing how inverses function in practical scenarios.
Other exercises in this chapter
Problem 2
Simplify each rational expression. State any restrictions on the variable. $$ \frac{6 c^{2}+9 c}{3 c} $$
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Find any points of discontinuity for each rational function. $$ y=\frac{x^{2}+2 x}{x^{2}+2} $$
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Suppose that \(x\) and \(y\) vary inversely. Write a function that models each inverse variation. $$ x=-13 \text { when } y=100 $$
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The focal length \(f\) of a camera lens is 2 in. The lens equation is \(\frac{1}{f}=\frac{1}{d_{i}}+\frac{1}{d_{o}}\) where \(d_{i}\) is the distance between th
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