Problem 26
Question
Find an equation of the line that passes through the given point and has the indicated slope. Sketch the line by hand. Use a graphing utility to verify your sketch, if possible. $$(-3,6), \quad m=-3$$
Step-by-Step Solution
Verified Answer
The equation of the line that goes through the point (-3,6) and has a slope of -3 is \(y = -3x - 3\)
1Step 1: Write the Point-Slope Form of the Linear Equation
The point-slope form of a linear equation is \[y - y_1 = m(x - x_1)\]. Here, \((x_1, y_1) = (-3,6)\) and the slope \(m = -3\)
2Step 2: Substitute the Given Values
Substituting \((x_1, y_1) = (-3,6)\) and \(m = -3\) into the equation gives \[y - 6 = -3(x - (-3))\]
3Step 3: Simplify the Equation
Upon simplifying, the equation becomes \[y - 6 = -3x - 9\], further simplification leads to the equation in slope-intercept form \[y = -3x - 3\]. This is the equation of the line.
4Step 4: Sketch the Line
To sketch this line on paper, begin by plotting the point (-3,6). The negative slope of -3 indicates that the line falls as it moves to the right. Starting from the point (-3,6), move 3 units down and one unit to the right three times as the slope -3 means 'fall 3' and 'run 1'. This will give a rough sketch of the line.
5Step 5: Verify Using a Graphing Utility
If possible, this sketch can be checked by inputing the equation \(y = -3x - 3\) into a graphing utility, which should show the line passing through the point (-3,6) with a slope of -3.
Key Concepts
Linear EquationsSlope-Intercept FormGraphing Linear Equations
Linear Equations
Linear equations are the simplest form of equations you'll come across in algebra. They describe a straight line when plotted on a graph. These equations typically look like this: \( y = mx + b \) or \( ax + by = c \) where \( m \) is the slope and \( b \) is the y-intercept – the point where the line crosses the y-axis.
In our exercise, we have the task of finding the equation of a line given a point it passes through, \( (-3,6) \) and its slope, \( m=-3 \). 'Linear' refers to the line produced when the equation is graphed, as it's straight regardless of its slope or intercepts. The beauty of these equations lies in their simplicity and the fact that they model many real-world situations, ranging from physics concepts like speed to financial models for cost projection.
In our exercise, we have the task of finding the equation of a line given a point it passes through, \( (-3,6) \) and its slope, \( m=-3 \). 'Linear' refers to the line produced when the equation is graphed, as it's straight regardless of its slope or intercepts. The beauty of these equations lies in their simplicity and the fact that they model many real-world situations, ranging from physics concepts like speed to financial models for cost projection.
Slope-Intercept Form
how steep the line is and where it intersects the y-axis.
In step 3 of our solution, we simplify the point-slope equation to the slope-intercept form \( y = -3x - 3 \) to easily identify these characteristics. The slope \( m = -3 \) tells us that for every unit the line travels horizontally to the right (increase in \( x \) ), it drops down 3 units vertically (decrease in \( y \) ). Similarly, the y-intercept \( b = -3 \) informs us that the line crosses the y-axis at \( y = -3 \) .
In step 3 of our solution, we simplify the point-slope equation to the slope-intercept form \( y = -3x - 3 \) to easily identify these characteristics. The slope \( m = -3 \) tells us that for every unit the line travels horizontally to the right (increase in \( x \) ), it drops down 3 units vertically (decrease in \( y \) ). Similarly, the y-intercept \( b = -3 \) informs us that the line crosses the y-axis at \( y = -3 \) .
Graphing Linear Equations
Lastly, let's discuss graphing linear equations. It's a visual way to represent solutions to linear equations and brings the abstract concepts of algebra into a more intuitive, visual form.
To graph a linear equation, you need two things: a point on the line and the slope. The point gives you a concrete location to start from, and the slope shows you how to move to draw the line. In step 4 of our solution, you start at the given point \( (-3,6) \) and use the slope of \( -3 \) to graph. The negative sign of the slope indicates that as you move to the right on the x-axis, the line will fall, tracing out the line's path.
After plotting the known point, you use the slope to find another point, a process often called 'rise over run'. A slope of \( -3 \) means you go down 3 (rise) and to the right 1 (run). By drawing this line, you've graphed the linear equation. If you've got access to a graphing utility, as mentioned in step 5, you can always check your work by inputting the linear equation and ensuring your hand-drawn line matches the digital one.
To graph a linear equation, you need two things: a point on the line and the slope. The point gives you a concrete location to start from, and the slope shows you how to move to draw the line. In step 4 of our solution, you start at the given point \( (-3,6) \) and use the slope of \( -3 \) to graph. The negative sign of the slope indicates that as you move to the right on the x-axis, the line will fall, tracing out the line's path.
After plotting the known point, you use the slope to find another point, a process often called 'rise over run'. A slope of \( -3 \) means you go down 3 (rise) and to the right 1 (run). By drawing this line, you've graphed the linear equation. If you've got access to a graphing utility, as mentioned in step 5, you can always check your work by inputting the linear equation and ensuring your hand-drawn line matches the digital one.
Other exercises in this chapter
Problem 26
Sketch the graph of the equation. \(y=\sqrt{1-x}\)
View solution Problem 26
Determine whether the equation represents \(y\) as a function of \(x.\) $$x+y^{2}=3$$
View solution Problem 27
Use a graphing utility to graph the function and (b) determine the open intervals on which the function is increasing, decreasing, or constant. $$f(x)=3$$
View solution Problem 27
Evaluate the indicated function for \(f(x)=x^{2}-1\) and \(g(x)=x-2\) algebraically. If possible, use a graphing utility to verify your answer. $$(f-g)(t+1)$$
View solution