Problem 39
Question
Write an equation for a direct variation with a graph that passes through each point. \((7,2)\)
Step-by-Step Solution
Verified Answer
The equation for the direct variation that passes through the point (7,2) is \(y = \frac{2}{7}x\).
1Step 1: Understanding Direct Variation
A direct variation is described by an equation in the form of \(y = kx\), where \(k\) is the constant of variation. The graph of this equation will be a straight line through the origin (0,0) and the given point.
2Step 2: Calculating the Constant of Variation
To find the constant of variation (\(k\)), use the coordinates of the given point \((7, 2)\). Substitute these coordinates into the direct variation equation \(y = kx\) to get \(2 = k(7)\). Solving for \(k\) gives us \(k = \frac{2}{7}\).
3Step 3: Writing the Final Equation
Now that we have calculated the constant of variation to be \(\frac{2}{7}\), we can write the equation for the direct variation as \(y = \frac{2}{7}x\).
Key Concepts
Constant of VariationLinear EquationsAlgebraic Expressions
Constant of Variation
Understanding the concept of a constant of variation is central to grasping direct variation equations. In any direct variation, such as the equation \(y = kx\), the symbol \(k\) stands for the constant of variation, which is essentially the slope of the line on a graph. This constant determines how the variables \(x\) and \(y\) are related. For every unit increase in \(x\), the value of \(y\) will increase by \(k\) units.
To determine the value of this constant from a point on the line, one can use the formula \(k = \frac{y}{x}\). This is exactly what was done in the provided example with point \((7, 2)\). By plugging these values into the formula, the constant of variation was found to be \(\frac{2}{7}\). When you understand this concept, you can not only write the equation but also predict how one variable will change in direct proportion to the other.
To determine the value of this constant from a point on the line, one can use the formula \(k = \frac{y}{x}\). This is exactly what was done in the provided example with point \((7, 2)\). By plugging these values into the formula, the constant of variation was found to be \(\frac{2}{7}\). When you understand this concept, you can not only write the equation but also predict how one variable will change in direct proportion to the other.
Linear Equations
Linear equations form the foundation of algebra and graphically represent straight lines. Any equation that can be written in the form \(y = mx + b\) is a linear equation, where \(m\) is the slope and \(b\) is the y-intercept. When an equation shows direct variation, like the one in the exercise, it means that the y-intercept is zero (\(b = 0\)), simplifying the equation to \(y = mx\) or \(y = kx\).
In the context of direct variation and the associated linear equation, the slope \(m\) is replaced by the constant of variation \(k\). This is because the line always passes through the origin, making the graph a perfect representation of proportional change. The key to understanding linear equations in direct variation scenarios is recognizing the role of the slope as a multiplier that defines the rate at which one variable changes with respect to the other.
In the context of direct variation and the associated linear equation, the slope \(m\) is replaced by the constant of variation \(k\). This is because the line always passes through the origin, making the graph a perfect representation of proportional change. The key to understanding linear equations in direct variation scenarios is recognizing the role of the slope as a multiplier that defines the rate at which one variable changes with respect to the other.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and operations that represent a specific value. They become especially useful when forming equations to model real-world situations or abstract concepts like direct variation. In our example, the algebraic expression \(y = \frac{2}{7}x\) succinctly states that \(y\) varies directly as \(x\).
The beauty of algebraic expressions lies in their ability to be manipulated according to algebraic principles. For instance, if we needed to find the value of \(y\) for a given \(x\), or vice versa, we could easily do so with our expression. It's important to be comfortable with rearranging these expressions, as this skill will often be employed to isolate variables, rewrite equations in different forms, and solve complex problems across various areas of mathematics and science.
The beauty of algebraic expressions lies in their ability to be manipulated according to algebraic principles. For instance, if we needed to find the value of \(y\) for a given \(x\), or vice versa, we could easily do so with our expression. It's important to be comfortable with rearranging these expressions, as this skill will often be employed to isolate variables, rewrite equations in different forms, and solve complex problems across various areas of mathematics and science.
Other exercises in this chapter
Problem 39
Find the range of each function when the domain is \(\\{-3,-1,0,1.5,4\\}\). $$ f(x)=2 x-1 $$
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Write an equation for each line. Then graph the line. through \((-3,-1)\) and perpendicular to \(y=-\frac{2}{5} x-4\)
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a. Graph the parent function \(f(x)=x\) and the function \(g(x)=3 x\) on a coordinate plane. b. Translate the second graph 5 units up. Write an equation for the
View solution Problem 40
Graph each absolute value equation. $$ y=-|x-3| $$
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