Chapter 2: Problem 16
Exer. 15-16: Sketch the graph of the line through \(P\) for each value of \(m\) $$P(-2,4) ; \quad m=1,-2,-\frac{1}{2}$$
Short Answer
Expert verified
Three lines: \(y = x + 6\), \(y = -2x\), \(y = -\frac{1}{2}x + 3\).
Step by step solution
01
Understanding the Slope-Point Form
The equation of a line in point-slope form is given by \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. Here, the point \(P\) is \((-2, 4)\).
02
Slope \(m = 1\)
Substitute the point \((-2, 4)\) and slope \(m = 1\) into the point-slope form: \[ y - 4 = 1(x + 2) \]. Simplify to find the equation of the line in slope-intercept form: \[ y = x + 6 \].
03
Sketch the Line for \(m = 1\)
For \(y = x + 6\), the line crosses the y-axis at \(6\) and has a slope of \(1\), meaning it rises one unit for each unit it moves to the right. Plot \(P(-2, 4)\) and draw a line with the slope passing through this point.
04
Slope \(m = -2\)
Substitute \(m = -2\) into the point-slope form: \[ y - 4 = -2(x + 2) \].Simplify to find: \[ y = -2x \].
05
Sketch the Line for \(m = -2\)
For \(y = -2x\), the line passes through the origin with a y-intercept of \(0\) and a slope of \(-2\), descending two units for every unit it moves to the right. Plot \(P(-2, 4)\) and draw the line through it with this slope.
06
Slope \(m = -\frac{1}{2}\)
Substitute \(m = -\frac{1}{2}\) into the point-slope form: \[ y - 4 = -\frac{1}{2}(x + 2) \].Simplify to find: \[ y = -\frac{1}{2}x + 3 \].
07
Sketch the Line for \(m = -\frac{1}{2}\)
For \(y = -\frac{1}{2}x + 3\), the line has a y-intercept at \(3\) and a slope of \(-\frac{1}{2}\), descending one-half unit for every one unit it moves right. Plot \(P(-2, 4)\), and draw the line using this slope.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope and Y-Intercept
Understanding the slope and y-intercept of a linear equation is crucial for graphing lines effectively. Let's start with the basics: the y-intercept is the point where the line crosses the y-axis. This point occurs when the x-value is zero. Hence, when you have a linear equation in slope-intercept form, which looks like this: \( y = mx + b \), the y-intercept is represented by \( b \). The slope, indicated by \( m \), tells you how steep the line is. It represents the rate of change, or how much the y-value of the line changes for each unit change in the x-value.
For example, in the equation \( y = x + 6 \), the y-intercept is \( 6 \), meaning the line crosses the y-axis at \( (0, 6) \). The slope is \( 1 \), indicating that for each step to the right in the x-direction, the line rises by one unit.
For example, in the equation \( y = x + 6 \), the y-intercept is \( 6 \), meaning the line crosses the y-axis at \( (0, 6) \). The slope is \( 1 \), indicating that for each step to the right in the x-direction, the line rises by one unit.
- Slope \( m = 1 \): line tilts upward.
- Y-Intercept \( b = 6 \): where the line crosses the y-axis.
Point-Slope Form
The point-slope form of a linear equation is useful when you know a point on the line and its slope. This form is particularly handy for quickly generating an equation when you have the coordinates of a point and the slope value. The point-slope equation is formulated as \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a specific point on the line, and \(m\) is the slope.
To illustrate, suppose you're given the point \( P(-2, 4) \) and a slope \( m = -2 \). Plugging these values into the point-slope form gives \( y - 4 = -2(x + 2) \). You can simplify this to get \( y = -2x \). This tells us that the line goes through the point \((-2, 4)\) and has a downward slope of \(-2\), moving down two units for every step it moves to the right.
To illustrate, suppose you're given the point \( P(-2, 4) \) and a slope \( m = -2 \). Plugging these values into the point-slope form gives \( y - 4 = -2(x + 2) \). You can simplify this to get \( y = -2x \). This tells us that the line goes through the point \((-2, 4)\) and has a downward slope of \(-2\), moving down two units for every step it moves to the right.
- Easy to use when starting with a specific point.
- Converts to slope-intercept form for easier graphing.
Linear Equations
Linear equations form the backbone of algebra, represented graphically as straight lines. These equations express a relationship where one variable depends additively on another. The standard approach for formulating a linear equation is to arrange it in the slope-intercept form: \( y = mx + b \). This format highlights how the equation relates to its graph.
Consider the different scenarios as slopes change, but the standard linear structure holds. Take \( y = -\frac{1}{2}x + 3 \) as an example. Here, \( -\frac{1}{2} \) is the slope, meaning the line decreases half a unit vertically for every unit it advances horizontally, and \( 3 \) is the y-intercept, where the line touches the y-axis.
Consider the different scenarios as slopes change, but the standard linear structure holds. Take \( y = -\frac{1}{2}x + 3 \) as an example. Here, \( -\frac{1}{2} \) is the slope, meaning the line decreases half a unit vertically for every unit it advances horizontally, and \( 3 \) is the y-intercept, where the line touches the y-axis.
- Linear directly denotes straightness.
- Expands to various forms like point-slope for certain tasks.