Problem 37
Question
Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. \(x\) -intercept \(=-\frac{1}{2}\) and \(y\) -intercept \(=4\)
Step-by-Step Solution
Verified Answer
The equation of the line in point-slope form is \(y - 4 = -8x\) and in slope-intercept form is \(y = -8x + 4\).
1Step 1: Find the slope of the line
A line is defined by its slope and intercept. We know that the slope of the line, \(m\), is given by the formula, \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For the line that passes through the x-intercept and y-intercept points, we have \(x_1=-\frac{1}{2}, y_1=0\) (x-intercept) and \(x_2=0, y_2=4\) (y-intercept). So the slope \(m = \frac{4-0}{0 - -\frac{1}{2}} = -8 \).
2Step 2: Write the equation of the line in point-slope form
The point-slope form of a line is, \(y - y_1 = m(x - x_1)\). Using the slope calculated in Step 1 and one of the points, say the y-intercept \(0,4\) , the equation of the line in point-slope form is \(y - 4 = -8(x - 0)\). So the equation in point-slope form is \(y - 4 = -8x \).
3Step 3: Write the equation of the line in slope-intercept form
The slope intercept form is, \(y = mx + b\), where \(b\) is the y-intercept. By substituting the slope calculated in step 1 and the y-intercept given in the problem into the slope-intercept equation, we get: \(y = -8x + 4\) .
Key Concepts
Point-Slope FormSlope-Intercept FormX-Intercept and Y-Intercept
Point-Slope Form
Understanding the point-slope form is crucial for writing equations of lines when given a point on the line and the slope. It is expressed as \(y - y_1 = m(x - x_1)\), where \(m\) represents the slope and \(x_1\), \(y_1\) are the coordinates of the given point through which the line passes.
To find the equation of a line using this form, you need one specific point on the line and the slope. For instance, if you're given the x-intercept, which is where the line crosses the x-axis, and the y-intercept, where the line crosses the y-axis, you can calculate the slope by finding the change in y over the change in x between these two points. Once you have the slope and a point, you can plug these values into the point-slope form to get the equation of the line.
Using the exercise, the slope we found is \(m = -8\), and taking the y-intercept as a point, \(x_1 = 0\), \(y_1 = 4\), we insert these into the formula to get \(y - 4 = -8(x - 0)\), which simplifies to \(y - 4 = -8x\). This method is particularly useful for visualizing how a line behaves in relation to a specific point, and it can be quickly transformed into other forms with simple algebraic manipulation.
To find the equation of a line using this form, you need one specific point on the line and the slope. For instance, if you're given the x-intercept, which is where the line crosses the x-axis, and the y-intercept, where the line crosses the y-axis, you can calculate the slope by finding the change in y over the change in x between these two points. Once you have the slope and a point, you can plug these values into the point-slope form to get the equation of the line.
Using the exercise, the slope we found is \(m = -8\), and taking the y-intercept as a point, \(x_1 = 0\), \(y_1 = 4\), we insert these into the formula to get \(y - 4 = -8(x - 0)\), which simplifies to \(y - 4 = -8x\). This method is particularly useful for visualizing how a line behaves in relation to a specific point, and it can be quickly transformed into other forms with simple algebraic manipulation.
Slope-Intercept Form
The slope-intercept form of a linear equation is possibly the most familiar to students: \(y = mx + b\). This equation provides a straightforward way to graph a line when you know the line's slope and y-intercept.
The symbol \(m\) represents the slope, and \(b\) represents the y-intercept, which is the point where the line crosses the y-axis (at \(x = 0\)). To write the equation of a line in slope-intercept form using the x-intercept and y-intercept, you first determine the slope (\(m\)) using the change in y over the change in x between these intercepts.
Continuing with our exercise example, after finding the slope to be \(m = -8\), and knowing the y-intercept (\(b\)) to be \(4\), we substitute these values into the slope-intercept form and obtain \(y = -8x + 4\). This form is highly preferred when graphing because it directly shows both the steepness and the initial value of the line, making it extremely user-friendly for quick sketches and interpretation.
The symbol \(m\) represents the slope, and \(b\) represents the y-intercept, which is the point where the line crosses the y-axis (at \(x = 0\)). To write the equation of a line in slope-intercept form using the x-intercept and y-intercept, you first determine the slope (\(m\)) using the change in y over the change in x between these intercepts.
Continuing with our exercise example, after finding the slope to be \(m = -8\), and knowing the y-intercept (\(b\)) to be \(4\), we substitute these values into the slope-intercept form and obtain \(y = -8x + 4\). This form is highly preferred when graphing because it directly shows both the steepness and the initial value of the line, making it extremely user-friendly for quick sketches and interpretation.
X-Intercept and Y-Intercept
The concepts of x-intercept and y-intercept are fundamental in understanding the behavior of linear equations on a coordinate plane. The x-intercept is the point where the graph of the equation crosses the x-axis, while the y-intercept is where it crosses the y-axis.
When we are looking to graph a line or write its equation, these intercepts can provide important clues. Typically, the x-intercept is found by setting \(y = 0\) and solving for \(x\), and vice versa for the y-intercept.
In our example, we have the x-intercept as \(x = -\frac{1}{2}\) and the y-intercept as \(y = 4\). They served as our starting point to find the slope by calculating the change in y over the change in x from the x-intercept to the y-intercept. These intercepts not only contribute to the slope calculation but are also used directly in the slope-intercept form of the line, which in our case leads to the final equation \(y = -8x + 4\). Recognizing and accurately plotting these intercepts are essential steps for graphing and analyzing linear equations.
When we are looking to graph a line or write its equation, these intercepts can provide important clues. Typically, the x-intercept is found by setting \(y = 0\) and solving for \(x\), and vice versa for the y-intercept.
In our example, we have the x-intercept as \(x = -\frac{1}{2}\) and the y-intercept as \(y = 4\). They served as our starting point to find the slope by calculating the change in y over the change in x from the x-intercept to the y-intercept. These intercepts not only contribute to the slope calculation but are also used directly in the slope-intercept form of the line, which in our case leads to the final equation \(y = -8x + 4\). Recognizing and accurately plotting these intercepts are essential steps for graphing and analyzing linear equations.
Other exercises in this chapter
Problem 36
Write the standard form of the equation of the circle with the given center and radius. $$\text { Center }(-3,5), r=3$$
View solution Problem 37
Begin by graphing the standard cubic function, \(f(x)=x^{3} .\) Then use transformations of this graph to graph the given function. $$ g(x)=(x-3)^{3} $$
View solution Problem 37
Find: a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=4-x^{2}, g(x)=\sqrt{x^{2}-4}$$
View solution Problem 37
In Exercises \(33-44\), find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$for the given function. $$f(x)=x^{2}$$
View solution