/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Draw the graph of each equation.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Draw the graph of each equation. Name any intercepts. $$\frac{1}{2} x+y=3$$

Short Answer

Expert verified
The y-intercept is (0, 3); the x-intercept is (6, 0).

Step by step solution

01

Rearrange the Equation

Start with the equation \(\frac{1}{2}x + y = 3\). To graph this, we need it in the slope-intercept form \(y = mx + b\). First, subtract \(\frac{1}{2}x\) from both sides:\[y = -\frac{1}{2}x + 3\] This form makes it easy to identify the slope \(-\frac{1}{2}\) and the y-intercept \(3\).
02

Identify the Y-Intercept

From the equation \(y = -\frac{1}{2}x + 3\), we can immediately see that the y-intercept is where \(x = 0\). Substitute \(x = 0\) into the equation:\[y = -\frac{1}{2}(0) + 3 = 3\] Therefore, the y-intercept is \((0, 3)\).
03

Identify the X-Intercept

To find the x-intercept, set \(y = 0\) in the equation and solve for \(x\):\[0 = -\frac{1}{2}x + 3\]\[\frac{1}{2}x = 3\]\[x = 6\]Therefore, the x-intercept is \((6, 0)\).
04

Plot the Intercepts and Draw the Line

Use the intercepts \((0, 3)\) and \((6, 0)\) to plot two points on the graph. Draw a straight line through these points to represent the line described by the equation \(\frac{1}{2}x + y = 3\). This is the graph of the equation.

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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 of a linear equation is represented as \( y = mx + b \). This form is particularly useful because it gives us immediate insights into two key components of a linear equation: the slope and the y-intercept.

The letter \( m \) indicates the slope of the line. The slope measures the "steepness" of the line, showing how much \( y \) changes for a given change in \( x \). If the slope is positive, the line inclines upwards; if negative, it declines downwards. For example, in the equation we worked with, \( y = -\frac{1}{2}x + 3 \), the slope is \(-\frac{1}{2}\). This means for every 2 units the line runs horizontally to the right, it falls 1 unit.
  • This equation was rearranged from its original form, \( \frac{1}{2}x + y = 3 \), to fit the slope-intercept form by isolating \( y \).
  • The constant \( b \) in the equation (\( b = 3 \)) indicates where the line crosses the y-axis.
By using this form, you can quickly draw a graph without needing to calculate many points, simplifying the graphing process dramatically.
Y-Intercept
The y-intercept of a line is the point where the line crosses the y-axis. This happens when the value of \( x \) is equal to zero.

To find the y-intercept, you simply look at the constant term \( b \) in the slope-intercept form \( y = mx + b \). In our example, the equation \( y = -\frac{1}{2}x + 3 \) tells us directly that the y-intercept is 3.
This means that when \( x = 0 \), \( y \) equals 3. Therefore, the coordinate is \( (0, 3) \).
  • The beauty of the y-intercept is that it's easily visible from the equation when in slope-intercept form.
  • It's also crucial for rapidly graphing a line, as it provides one of the two points needed to draw a straight line on a graph.
X-Intercept
The x-intercept is the point where the line crosses the x-axis, which occurs when \( y = 0 \).

To find the x-intercept from the equation \( y = -\frac{1}{2}x + 3 \), we set \( y \) to 0 and solve for \( x \):
  • Start by reorganizing the equation: \( 0 = -\frac{1}{2}x + 3 \).
  • Solving gives you \( \frac{1}{2}x = 3 \) by adding \( \frac{1}{2}x \) to both sides.
  • Finally, multiply both sides by 2 to find \( x = 6 \).
This tells us that when the line hits the x-axis, it does so at the point \( (6, 0) \). Finding the x-intercept is crucial for completing the graph.
Using both x- and y-intercepts, you have two points which are sufficient to draw the linear graph accurately.

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