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Chapter 0: Problem 3

Graph the following equations. $$y=3 x+1$$

Short Answer

Expert verified
Plot points (0, 1) and (1, 4) then draw the line through them.

Step by step solution

01

- Identify the equation type

The given equation is in slope-intercept form which is represented as \(y = mx + b\). Here, \(m\) denotes the slope and \(b\) represents the y-intercept.
02

- Determine the slope and y-intercept

From the equation \( y = 3x + 1 \), identify the slope \(m\) as 3 and the y-intercept \(b\) as 1.
03

- Plot the y-intercept

On a graph, locate the point where the line crosses the y-axis. This is the y-intercept (0, 1). Place a point at (0, 1) on the graph.
04

- Use the slope to find the next point

The slope of 3 indicates rising 3 units for every 1 unit it runs to the right. From the y-intercept (0, 1), move 1 unit right to (1, 1) and then 3 units up to (1, 4). Place a point at (1, 4).
05

- Draw the line

Draw a straight line through the points (0, 1) and (1, 4) extending it in both directions. This line is the graph of the equation \(y = 3x + 1\).

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Key Concepts

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

slope-intercept form
The slope-intercept form is a common way to write a linear equation. It is written as y = mx + bwhere:
  • 'm' is the slope of the line, which tells us how steep the line is.
  • 'b' is the y-intercept, the point where the line crosses the y-axis.
Using this form makes graphing linear equations easier because it directly shows the slope and y-intercept.
slope
The slope of a line ( 'm' ) indicates the steepness and direction of the line. It is calculated as the rise (change in y) over the run (change in x). For our example with the equation y = 3x + 1, the slope is 3. This means that for every 1 unit moved to the right along the x-axis, the line rises by 3 units on the y-axis. If the slope were negative, the line would fall instead of rising. Remember, a slope of 0 means the line is horizontal, and an undefined slope means the line is vertical.
y-intercept
The y-intercept ( 'b' ) is the point where the line crosses the y-axis. In the equation y = 3x + 1, the y-intercept is 1. This means that when x = 0, y = 1. To plot this on a graph, you would find the point (0, 1). This is the starting point for drawing your line. No matter the slope, every line in slope-intercept form will cross the y-axis at the y-intercept value.
plotting points
Plotting points is essential when graphing a linear equation. After identifying the y-intercept '(0, 1)' in our example:
  • Start by placing a point at (0, 1) on the graph.
  • Next, use the slope to find another point.
For the equation y = 3x + 1, the slope is 3. This tells us to move 1 unit right and 3 units up from our initial point, leading to the point (1, 4). Once you have two points, you can sketch the straight line that represents the equation.
linear equation
A linear equation is an equation that forms a straight line when graphed. It can be written in various forms, including the slope-intercept form (y = mx + b). It shows a constant rate of change, meaning the relationship between x and y is consistent. In our example, the equation y = 3x + 1 represents a linear relationship. As x changes, y changes consistently according to the slope. Linear equations are foundational in algebra and help understand relationships between variables.

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Most popular questions from this chapter

Shifting a Graph Let \(f(x)=x^{2} .\) Graph the functions \(f(x)+1\) \(f(x)-1, f(x)+2,\) and \(f(x)-2 .\) Make a guess about the relationship between the graph of a general function \(f(x)\) and the graph of \(f(x)+c\) for some constant \(c .\) Test your guess on the functions \(f(x)=x^{3}\) and \(f(x)=\sqrt{x}\).

Let \(f(x)=\frac{x}{x-2}, g(x)=\frac{5-x}{5+x},\) and \(h(x)=\frac{x+1}{3 x-1} .\) Express the following as rational functions. $$\frac{g(x+5)}{f(x+5)}$$

Find the zeros of the function. (Use the specified viewing window.) $$f(x)=\frac{x}{x+2}-x^{2}+1 ;[-1.5,2] \text { by }[-2,3]$$

Let \(f(x)=x^{6}, g(x)=\frac{x}{1-x},\) and \(h(x)=x^{3}-5 x^{2}+1 .\) Calculate the following functions. $$h(f(t))$$

Use the laws of exponents to simplify the algebraic expressions. Your answer should not involve parentheses or negative exponents. $$\frac{x^{3}}{y^{-2}}$$
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