Problem 15
Question
Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The horizontal line through \(\left(4, \frac{1}{4}\right)\)
Step-by-Step Solution
Verified Answer
The standard form is \(4y = 1\).
1Step 1: Identify Characteristics of a Horizontal Line
A horizontal line has a slope of 0. This means its equation in slope-intercept form, \(y = mx + b\), simplifies to \(y = b\) because the slope \(m = 0\). All points on the line have the same \(y\)-coordinate.
2Step 2: Use the Given Point
The given point is \( (4, \frac{1}{4}) \). Since the line is horizontal, every point on the line will have the same \(y\)-coordinate, including the given point. Thus, we know that \(y = \frac{1}{4}\).
3Step 3: Convert the Equation to Standard Form
The goal is to convert the equation \(y = \frac{1}{4}\) to standard form. The standard form of a line is \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) should be positive. Start by writing \(0\cdot x + 1\cdot y = \frac{1}{4}\), which simplifies to \(y = \frac{1}{4}\).
4Step 4: Eliminate the Fraction
To eliminate the fraction in the standard form, multiply each term by 4 to make the \(y\)-term an integer: \[4(0)x + 4(1)y = 4\left(\frac{1}{4}\right)\]. Simplifying gives \(0x + 4y = 1\) or just \(4y = 1\).
5Step 5: Conclude the Standard Form Equation
The equation in standard form is \(0x + 4y = 1\), which simplifies to \(4y = 1\). This format satisfies the required conditions, making \(A = 0\), \(B = 4\), and \(C = 1\).
Key Concepts
Horizontal LineSlope-Intercept FormEquation Conversion
Horizontal Line
A horizontal line is a special type of straight line with unique characteristics. It runs parallel to the x-axis on a coordinate plane, meaning all points on the line have the same y-coordinate. This uniformity leads to a slope of zero, making it much easier to determine its equation.
It’s important to note that the equation of a horizontal line can be expressed as simply as \( y = b \), where \( b \) is the constant y-coordinate shared by all points on the line. For instance, if we know the line passes through the point \( (4, \frac{1}{4}) \), the y-coordinate is \( \frac{1}{4} \), so the line's equation in slope-intercept form is \( y = \frac{1}{4} \).
It’s important to note that the equation of a horizontal line can be expressed as simply as \( y = b \), where \( b \) is the constant y-coordinate shared by all points on the line. For instance, if we know the line passes through the point \( (4, \frac{1}{4}) \), the y-coordinate is \( \frac{1}{4} \), so the line's equation in slope-intercept form is \( y = \frac{1}{4} \).
- Horizontal lines are constant in the y-value.
- Viewer easily sees them as flat and level.
- They stretch across the graph without tilting up or down.
Slope-Intercept Form
Understanding the slope-intercept form of a line's equation is a foundational concept in algebra. This form is expressed as \( y = mx + b \), where \( m \) represents the slope of the line, and \( b \) is the y-intercept. This format is useful because it immediately tells the slope and starting value (intercept) of the line.
Given that horizontal lines have a slope \( m = 0 \), the equation simplifies to \( y = b \). This straightforward result lets you describe any horizontal line by simply identifying its constant y-value, thereby eliminating the need to calculate additional intercepts or listings.
Given that horizontal lines have a slope \( m = 0 \), the equation simplifies to \( y = b \). This straightforward result lets you describe any horizontal line by simply identifying its constant y-value, thereby eliminating the need to calculate additional intercepts or listings.
- Slope \( m \) provides line steepness, but for horizontal, it is 0.
- Y-intercept \( b \) tells you where the line crosses the y-axis.
Equation Conversion
Converting an equation from one form to another can enhance our understanding and make solving problems easier. When working with equations of lines, transitioning from the slope-intercept form \( y = mx + b \) to the standard form \( Ax + By = C \) is a common task.
For horizontal lines, since the slope is 0, the equation \( y = \frac{1}{4} \) needs conversion to its standard form, where constants \( A, B, C \) should ideally be integers. This might involve multiplying the entire equation to clear any fractions.
For horizontal lines, since the slope is 0, the equation \( y = \frac{1}{4} \) needs conversion to its standard form, where constants \( A, B, C \) should ideally be integers. This might involve multiplying the entire equation to clear any fractions.
- The equation \( y = \frac{1}{4} \) converts to \( 4y = 1 \) in standard form.
- Standard form requires equations such that \( A \) and \( B \) are integers, with \( A \) non-negative.
Other exercises in this chapter
Problem 15
sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=\ln (x+1) $$
View solution Problem 15
Suppose that \(f(x)=1-x, x \in \mathbf{R}\), and \(g(x)=\sqrt{x}, x \geq 0\). (a) Find \((f \circ g)(x)\) together with its domain. (b) Find \((g \circ f)(x)\)
View solution Problem 16
sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=\ln (x-3) $$
View solution Problem 16
Suppose that \(f(x)=\frac{1}{x+1}, x \neq-1\), and \(g(x)=2 x^{2}, x \in \mathbf{R}\). (a) Find \((f \circ g)(x)\). (b) Find \((g \circ f)(x)\). In both (a) and
View solution