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Chapter 1: Problem 8

Change the equation \(2 x+3 y+6=0\) to the form \(y=m x+b\), graph the line, and find the \(y\) -intercept and \(x\) -intercept. \(\Rightarrow\)

Short Answer

Expert verified
Equation in form: \(y = -\frac{2}{3}x - 2\), \(y\)-intercept: (0, -2), \(x\)-intercept: (-3, 0).

Step by step solution

01

Re-arranging the Equation

The equation is given as \(2x + 3y + 6 = 0\). Our goal is to express it in the form \(y = mx + b\). Start by isolating the term with \(y\) on one side. Subtract \(2x\) and \(6\) from both sides to get \(3y = -2x - 6\).
02

Solving for y

Now, divide every term by 3 to solve for \(y\). This gives us \(y = -\frac{2}{3}x - 2\). Now the equation is in the desired slope-intercept form \(y = mx + b\) where \(m = -\frac{2}{3}\) and \(b = -2\).
03

Graphing the Line

To graph the line, you can start by plotting the intercepts. The \(y\)-intercept is \(b = -2\) which means the line crosses the \(y\)-axis at (0, -2). The slope \(m = -\frac{2}{3}\) indicates that for every 3 units you move to the right on the \(x\)-axis, you move 2 units down on the \(y\)-axis.
04

Finding the y-intercept

The \(y\)-intercept is the point where the line crosses the \(y\)-axis. For our equation, this is directly given by \(b = -2\). Thus, the \(y\)-intercept is at the point (0, -2).
05

Finding the x-intercept

To find the \(x\)-intercept, set \(y = 0\) in the equation \(2x + 3y + 6 = 0\). Solving for \(x\) gives \(2x + 6 = 0\) or \(2x = -6\). Dividing both sides by 2, we find \(x = -3\). Therefore, the \(x\)-intercept is at the point (-3, 0).

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Key Concepts

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

Slope-Intercept Form
Linear equations can often be expressed in different forms to highlight useful properties. One of these forms is the slope-intercept form, which looks like this: \( y = mx + b \). Here, \( m \) represents the **slope** of the line, and \( b \) is the **y-intercept**.

This form is particularly useful because it instantly tells you how steep the line is and where it crosses the y-axis. It makes graphing much easier too! In our original equation \( 2x + 3y + 6 = 0 \), the first step was to rearrange it to look like \( y = mx + b \). This gives us a clearer picture of the line's behavior on a graph.
Graph of a Line
Graphing a line from its equation gives a visual representation of all solutions to the equation. When the equation is in slope-intercept form, \( y = mx + b \), graphing is fairly straightforward.

To graph our line, we identified the y-intercept \( b = -2 \), meaning the line passes through the point (0, -2) on the y-axis. Additionally, the slope \( m = -\frac{2}{3} \) indicates the line's slant.

To plot the line,:
  • Start at the y-intercept (0, -2).
  • Use the slope: move 3 units to the right and 2 units down to find a second point.
  • Draw a line through these points.
The slope tells us the line's inclination, helping in creating an accurate graph.
Y-Intercept
The y-intercept is an essential characteristic of a line. This is where the line crosses the y-axis, and it's always at the point (0, b).

For our equation, once it was placed into slope-intercept form as \( y = -\frac{2}{3}x - 2 \), it was simple to pinpoint the y-intercept. Here, \( b = -2 \), telling us the line crosses the y-axis at the point (0, -2).

This direct reading of the intercept from the equation makes calculations straightforward, allowing us to graph the point on a graph quickly.
X-Intercept
The x-intercept is where the line crosses the x-axis. To find this point, set \( y = 0 \) and solve the equation for \( x \).

In our case, from the original equation \( 2x + 3y + 6 = 0 \), setting \( y = 0 \) resulted in \( 2x + 6 = 0 \). Solving this gives \( x = -3 \), indicating the line crosses the x-axis at the point (-3, 0).

Understanding both x- and y-intercepts enhances graphing, ensuring accuracy for visualizing linear equations.

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Most popular questions from this chapter

Starting with the graph of \(y=\sqrt{x},\) the graph of \(y=1 / x,\) and the graph of \(y=\sqrt{1-x^{2}}\) (the upper unit semicircle), sketch the graph of each of the following functions: $$f(x)=-1-1 /(x+2)$$

For each pair of points \(A\left(x_{1}, y_{1}\right)\) and \(B\left(x_{2}, y_{2}\right)\) find (i) \(\Delta x\) and \(\Delta y\) in going from \(A\) to \(B\), (ii) the slope of the line joining \(A\) and \(B,\) (iii) the equation of the line joining \(A\) and \(B\) in the form \(y=m x+b\), (iv) the distance from \(A\) to \(B\), and (v) an equation of the circle with center at \(A\) that goes through \(B\). a) \(A(2,0), B(4,3)\) b) \(A(1,-1), B(0,2)\) c) \(A(0,0), B(-2,-2)\) d) \(A(-2,3), B(4,3)\) e) \(A(-3,-2), B(0,0)\) f) \(A(0.01,-0.01), B(-0.01,0.05)\)

Find the domain of each of the following functions: $$y=f(x)=\sqrt{1-(1 / x)}=$$

Find the domain of each of the following functions: $$y=f(x)=\sqrt[3]{x}$$

Find the domain of each of the following functions: $$y=f(x)=1 /(\sqrt{x}-1)$$
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