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Chapter 2: Problem 70

Use intercepts to graph each equation. $$ 3 x+5 y+15=0 $$

Short Answer

Expert verified
The x-intercept is -5, and the y-intercept is -3.

Step by step solution

01

Find the x-intercept

Firstly, it's necessary to find the x-intercept. This is the point where the graph intersects the x-axis. At this point, \(y = 0.\) Set \(y = 0\) in the equation \(3x+5y+15=0\), and solve for \(x\). That gives us \(3x = -15\). Divide each side by \(3\) to find \(x = -5\). So, the x-intercept is \(-5\).
02

Find the y-intercept

Next, let's find the y-intercept, which is the value of \(y\) when \(x = 0.\) Put \(x=0\) in the equation \(3x+5y+15=0\), and solve for \(y\). It gives us \(5y = -15\). Divide each side by \(5\) to get \(y = -3\). So, the y-intercept is \(-3\).
03

Graph the Equation

Now, it's time to plot these intercepts on a graph and draw a line through them. Mark the points \((-5,0)\) and \((0,-3)\) on the x and y axis respectively. Draw a straight line through these points. This line represents the graph of the equation \(3x+5y+15=0\).

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Key Concepts

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

Intercepts
To successfully graph linear equations, understanding the concept of intercepts is crucial. An intercept is a point where the graph of a line crosses either the x-axis or the y-axis. In linear equations, these points are fundamental because they provide fixed locations through which the line must pass, making graphing much easier.

For any linear equation written in the form of \[ ax + by + c = 0, \]there are typically two intercepts to find:
  • X-intercept: The x-coordinate of the point where the line crosses the x-axis (where y=0).
  • Y-intercept: The y-coordinate of the point where the line crosses the y-axis (where x=0).
By identifying these two intercepts, one can quickly sketch the graph of the line by simply plotting these two points and drawing a line through them.
X-intercept
The x-intercept of a linear equation is fundamental in graphing the line. It signifies the point where the line meets the x-axis. At this point, the value of y is always zero.

To find the x-intercept from an equation such as \[ 3x + 5y + 15 = 0, \]you need to replace y with 0 and solve for x:
  • Set \( y = 0 \) in the equation which simplifies the equation to \( 3x + 15 = 0 \).
  • Then, solve for x by subtracting 15 from both sides to get \( 3x = -15 \).
  • Finally, divide both sides by 3 to find \( x = -5 \).
Therefore, the x-intercept is the point \((-5, 0)\). This single point tells where the graph touches the x-axis, and aids in determining the orientation of the line on the graph.
Y-intercept
Finding the y-intercept in a linear equation is integral to graphing as it shows where the line crosses the y-axis. At the y-intercept, the value of x will always be zero.

To determine the y-intercept using the equation \[ 3x + 5y + 15 = 0, \]follow these steps:
  • Set \( x = 0 \) so that the equation simplifies to \( 5y + 15 = 0 \).
  • Next, solve for y by subtracting 15 from both sides, resulting in \( 5y = -15 \).
  • Then, divide both sides by 5 to solve for y, finding \( y = -3 \).
Thus, the y-intercept is located at the point \((0, -3)\). This coordinate helps in setting up the graph because it shows a clear starting point on the y-axis, allowing for an easy visual plot in tandem with the x-intercept.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. Find the area of the donut-shaped region bounded by the graphs of \((x-2)^{2}+(y+3)^{2}=25\) and \((x-2)^{2}+(y+3)^{2}=36\)

What is a circle? Without using variables, describe how the definition of a circle can be used to obtain a form of its equation.

In your own words, describe how to find the distance between two points in the rectangular coordinate system.

Explaining the Concepts Does \((x-3)^{2}+(y-5)^{2}=-25\) represent the equation of a circle? What sort of set is the graph of this equation?

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=x^{2}+2, g(x)=x^{2}-2$$
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