Problem 42
Question
Give the slope and y-intercept of each line whose equation is given. Then graph the line. $$y=-3 x+2$$
Step-by-Step Solution
Verified Answer
The slope of the line is -3 and the y-intercept is 2. The graph of the line passes through the points (0,2) and (1,-1), slanting downwards.
1Step 1: Identify the Slope and the Y-Intercept
In the given equation \(y = -3x + 2\), \(m = -3\) is the slope and \(b = 2\) is the y-intercept.
2Step 2: Plot the Y-intercept
First step in drawing the graph is to plot the y-intercept on the y-axis. For this equation, plot a point at the coordinates (0,2) on the y-axis which corresponds to the y-intercept of 2.
3Step 3: Use the Slope to Find Another Point
Next, take the slope, or -3, which means move 3 units down for every 1 unit moved to the right from the point (0,2). So from the point (0,2), move 1 unit to the right and 3 units downwards. This new point is (1,-1). Plot this second point.
4Step 4: Draw the Line
Now that there are two points, (0,2) and (1,-1), a straight line can be drawn through them. This line represents the equation \(y = -3x + 2\).
Key Concepts
SlopeY-interceptPlotting PointsLinear Graphs
Slope
Understanding the slope of a line is crucial when dealing with linear equations. The slope is a numerical measure of a line's steepness and direction. It is often represented by the letter \(m\). In the equation \(y = -3x + 2\), the slope \(m\) is \-3\. This means for each unit you move to the right along the x-axis, you'll move 3 units down because of the negative sign, indicating that the line is descending from left to right.
When graphing, you can visualize the slope as a ratio of the vertical change (rise) to the horizontal change (run). In simpler terms, \(m = \frac{rise}{run}\). Slope can be positive (rising line), negative (falling line), zero (horizontal line), or undefined (vertical line). In our exercise, the negative slope suggests a downhill line from left to right on the graph.
When graphing, you can visualize the slope as a ratio of the vertical change (rise) to the horizontal change (run). In simpler terms, \(m = \frac{rise}{run}\). Slope can be positive (rising line), negative (falling line), zero (horizontal line), or undefined (vertical line). In our exercise, the negative slope suggests a downhill line from left to right on the graph.
Y-intercept
The y-intercept is another fundamental concept in graphing linear equations. It represents the point where the line crosses the y-axis. Expressed as \(b\) in the slope-intercept form \(y = mx + b\), it tells us the exact location on the y-axis where the line begins if you're plotting from left to right. For the equation \(y = -3x + 2\), the y-intercept is \(b = 2\), which corresponds to the point \(0, 2\) on the graph.
Plotting the y-intercept should always be your first step when drawing a line because it serves as an anchor point for the graph. From there, you can use the slope to determine the direction and steepness of the line, ensuring your graph will be accurate.
Plotting the y-intercept should always be your first step when drawing a line because it serves as an anchor point for the graph. From there, you can use the slope to determine the direction and steepness of the line, ensuring your graph will be accurate.
Plotting Points
Plotting points effectively translates algebraic equations into visual representations. This process begins with locating the y-intercept on the graph, as illustrated with our example \(y = -3x + 2\), and then plotting the point at \(0, 2\).
Subsequently, the slope \(m = -3\) informs us on how to find additional points. Starting from the y-intercept \(0, 2\), we move right 1 unit (run) and down 3 units (rise) — due to the negative slope — to locate the second point at \(1, -1\). With these two points, we can plot our line. It's important to remember that accuracy in this step leads to an accurate representation of the linear equation.
Subsequently, the slope \(m = -3\) informs us on how to find additional points. Starting from the y-intercept \(0, 2\), we move right 1 unit (run) and down 3 units (rise) — due to the negative slope — to locate the second point at \(1, -1\). With these two points, we can plot our line. It's important to remember that accuracy in this step leads to an accurate representation of the linear equation.
Linear Graphs
Linear graphs represent equations of the first degree. These graphs are straight lines, meaning their rate of change is constant, which is why the slope of these graphs is always the same at any two points along the line.
In the case of \(y = -3x + 2\), graphing it results in a straight line with a slope of \-3\ and a y-intercept at 2. To check the accuracy of a linear graph, any two points on the line can be used to verify if the slope, \(m\), remains constant. If additional points fall in a straight line with the initially plotted ones, you've confirmed the graph’s linearity and thus, the consistent application of the slope and y-intercept throughout the graph.
In the case of \(y = -3x + 2\), graphing it results in a straight line with a slope of \-3\ and a y-intercept at 2. To check the accuracy of a linear graph, any two points on the line can be used to verify if the slope, \(m\), remains constant. If additional points fall in a straight line with the initially plotted ones, you've confirmed the graph’s linearity and thus, the consistent application of the slope and y-intercept throughout the graph.
Other exercises in this chapter
Problem 41
Give the center and radius of the circle described by the equation and graph each equation. $$ x^{2}+y^{2}=16 $$
View solution Problem 42
Begin by graphing the standard cubic function, \(f(x)=x^{3} .\) Then use transformations of this graph to graph the given function. $$ h(x)=\frac{1}{4} x^{3} $$
View solution Problem 42
Express the given function h as a composition of two functions f and g so that \(h(x)=(f \circ g)(x)\) $$h(x)=\sqrt{5 x^{2}+3}$$
View solution Problem 42
In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=7$$
View solution