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Chapter 3: Problem 33

Exer. 33-36: Find the slope-intercept form of the line that satisfies the given conditions. $$ x \text {-intercept } 4, \quad y \text {-intercept }-3 $$

Short Answer

Expert verified
The slope-intercept form of the line is \( y = \frac{3}{4}x - 3 \).

Step by step solution

01

Understand Slope-Intercept Form

The slope-intercept form of a line is given by the equation \( y = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept.
02

Identify Intercepts

We are given that the x-intercept is 4 and the y-intercept is -3. At the x-intercept, the y-coordinate is 0, so the point is (4, 0). At the y-intercept, the x-coordinate is 0, so the point is (0, -3).
03

Calculate the Slope

The slope \( m \) is calculated using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Using our points (4, 0) and (0, -3), we find \( m = \frac{0 - (-3)}{4 - 0} = \frac{3}{4} \).
04

Formulate the Equation

Now that we have the y-intercept \( b = -3 \) and the slope \( m = \frac{3}{4} \), we can write the equation in slope-intercept form: \( y = \frac{3}{4}x - 3 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope Calculation
To find the slope of a line, we'll use two given points: the x-intercept and y-intercept. The slope, often represented by the letter "m", tells us how steep the line is. It's calculated by finding the change in y-values divided by the change in x-values from two points on the line.
The formula for calculating the slope is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Here:
  • \( (x_1, y_1) \) is the first point, or the x-intercept \((4, 0)\).
  • \( (x_2, y_2) \) is the second point, or the y-intercept \((0, -3)\).
Plugging these into the formula gives: \( m = \frac{0 - (-3)}{4 - 0} = \frac{3}{4} \).
This means that for every 4 units the line moves horizontally, it moves up 3 units vertically. This calculation is crucial to understanding the behavior and direction of the line in a graph.
X-Intercept
The x-intercept of a line is where the line crosses the x-axis. At this point, the y-value is 0. In our exercise, the x-intercept is given as 4.
This means the point (4, 0) lies on the line. Knowing the x-intercept can tell us where the line touches or crosses the horizontal axis of a graph. It's an essential part of defining the line and helps in graphing the equation, since it gives you a starting point on the x-axis from which you can plot the slope.
Y-Intercept
The y-intercept is where the line crosses the y-axis, and at this point, the x-value is 0. In our specific problem, the y-intercept is -3.
That means the line passes through the point (0, -3) on the graph. The y-intercept is conveniently used in the slope-intercept equation of a line \( y = mx + b \), where "b" is the y-intercept. It helps us easily start plotting when you have the equation of the line.
Knowing the y-intercept gives a clear picture of where the line started if you think of graphing it from the vertical axis, and it simplifies the process of formulating the entire equation.
Equation of a Line
To form the equation of a line, we use the slope-intercept form: \( y = mx + b \). This equation makes it simple to graph and understand lines by incorporating both the slope \( m \) and y-intercept \( b \).
From our calculations:
  • The slope \( m \) is \( \frac{3}{4} \), indicating the line rises 3 units for every 4 horizontal units.
  • The y-intercept \( b \) is -3, showing where the line crosses the y-axis.
Using these values, the equation becomes \( y = \frac{3}{4}x - 3 \). This tells us exactly how any point on the line relates to both x and y in terms of the rise over run (the slope) and the initial position on the y-axis (the y-intercept).

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Most popular questions from this chapter

If \(f\) is an odd function and \(g\) is an even function, is \(f g\) even, odd, or neither even nor odd?

Exer. 59-62: Sketch the graph of the equation. $$ y=|| x|-1| $$

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=\sqrt{c x}-1 ; \quad c=-1, \frac{1}{9}, 4 $$

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{x}{x-2}, \quad g(x)=\frac{3}{x} $$

Exer. 33-40: Explain how the graph of the function compares to the graph of \(y=f(x)\). For example, for the equation \(y=2 f(x+3)\), the graph of \(f\) is shifted 3 units to the left and stretched vertically by a factor of 2 . $$ y=-2 f\left(\frac{1}{3} x\right) $$
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