Problem 55
Question
Graph each relation or equation and find the domain and range. Then determine whether the relation or equation is a function. $$ x=-5 y+2 $$
Step-by-Step Solution
Verified Answer
The domain and range are both all real numbers, and the equation represents a function.
1Step 1: Rewrite the Equation
To graph the equation, we need to express it in terms of two variables that are easier to plot. Let's rewrite the given relation: \(x = -5y + 2\). We can solve for \(y\) in terms of \(x\) by rearranging it as \(5y = -x + 2\), thus \(y = \frac{-1}{5}x + \frac{2}{5}\). Now, it's easier to graph as a linear equation.
2Step 2: Graph the Equation
The equation \(y = \frac{-1}{5}x + \frac{2}{5}\) is in the form of \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. Here, \(m = \frac{-1}{5}\) and \(b = \frac{2}{5}\). Begin by plotting the y-intercept, which is \(\frac{2}{5}\), on the y-axis. From this point, use the slope \(\frac{-1}{5}\), which means moving down 1 unit for every 5 units you move to the right. Draw the line extending in both directions.
3Step 3: Find the Domain and Range
Since the line is continuous and infinite in both directions, the domain and range are both all real numbers. Therefore, the domain is \((-\infty, \infty)\) and the range is also \((-\infty, \infty)\).
4Step 4: Determine if it is a Function
A relation is a function if and only if each input (or x-value) is paired with exactly one output (or y-value). In this case, for each value of \(x\), there is exactly one corresponding value of \(y\). Hence, the relation is a function.
Key Concepts
Domain and RangeGraphing Linear EquationsSlope-Intercept Form
Domain and Range
When we talk about domain and range, we're dealing with the set of all possible values that a function can accept (domain) and produce (range).
Think of the domain as all possible inputs or x-values a function can take. For functions like the linear equation given, the domain can often include all real numbers, signifying inputs can be anything from negative infinity to positive infinity.
Just like with the domain, a linear equation extending infinitely in both directions allows the range to encompass all real numbers.
Think of the domain as all possible inputs or x-values a function can take. For functions like the linear equation given, the domain can often include all real numbers, signifying inputs can be anything from negative infinity to positive infinity.
- Domain: For our linear equation, since there's no constraint on x, the domain is (−∞, ∞).
Just like with the domain, a linear equation extending infinitely in both directions allows the range to encompass all real numbers.
- Range: This means for the equation, the range is also (−∞, ∞).
Graphing Linear Equations
Graphing linear equations involves plotting straight lines on a coordinate plane. A linear equation forms a line because it has variables raised only to the first power.
The standard way of graphing such an equation is by using the slope-intercept form, which we'll discuss in detail later.
The standard way of graphing such an equation is by using the slope-intercept form, which we'll discuss in detail later.
- Identify Intercepts: Identify the y-intercept from the equation as a starting point, such as the point (0, b).
- Use the Slope: From the y-intercept, apply the slope to locate another point on the line. For example, if the slope is −1/5, go down 1 unit and right 5 units to find another point.
- Draw the Line: Connect these points with a straight line. The graph represents all solutions to the equation.
Slope-Intercept Form
The slope-intercept form of a linear equation is given by the expression \( y = mx + b \). It's popular because it's straightforward and tells us so much about a line quickly.
- The Slope \( m \): Indicates the steepness or tilt of the line. A positive slope means the line ascends as x increases, while a negative slope implies descent. For our function, the slope is −1/5, suggesting the line slightly descends as it moves to the right.
- The Intercept \( b \): Represents the point where the line crosses the y-axis. Here, it is 2/5. This is often the first point plotted when graphing.
Other exercises in this chapter
Problem 55
PREREQUISITE SKILL. Solve each system of equations. $$ \begin{array}{l}{x+y=10} \\ {2 x+y=11}\end{array} $$
View solution Problem 55
State the dimensions of each matrix. $$ \left[\begin{array}{rrr}{7} & {5} & {0} \\ {-8} & {3} & {8} \\ {9} & {-1} & {15} \\ {4} & {2} & {11}\end{array}\right] $
View solution Problem 55
Find the \(x\)-intercept and the \(y\)-intercept of the graph of each equation. Then graph the equation. \(x-\frac{1}{2} y=8\)
View solution Problem 56
Evaluate each determinant. $$ \left|\begin{array}{rrr}{5} & {-7} & {3} \\ {-1} & {-2} & {-9} \\ {5} & {-7} & {3}\end{array}\right| $$
View solution