Graph the linear equation y=−3x+2

Question

Graph the linear equation y=−3x+2

.(1 point)
Responses

An illustration of a coordinate plane with four quadrants shows the x-axis ranging from negative 9 to 9 and the y-axis ranging from negative 11 to 11 in increments of 1. A solid downward line with arrows at both ends is plotted on the graph. he line passes through the plotted points left parenthesis 0 comma negative 2 right parenthesis and left parenthesis 1 comma negative 5 right parenthesis. A dotted vertical line labeled 3 from left parenthesis 1 comma negative 5 right parenthesis and a dotted horizontal line labeled 1 from left parenthesis 0 comma negative 2 right parenthesis intersect at left parenthesis 1 comma negative 2 right parenthesis, forming a right triangle with the slope.
Image with alt text: An illustration of a coordinate plane with four quadrants shows the x-axis ranging from negative 9 to 9 and the y-axis ranging from negative 11 to 11 in increments of 1. A solid downward line with arrows at both ends is plotted on the graph. he line passes through the plotted points left parenthesis 0 comma negative 2 right parenthesis and left parenthesis 1 comma negative 5 right parenthesis. A dotted vertical line labeled 3 from left parenthesis 1 comma negative 5 right parenthesis and a dotted horizontal line labeled 1 from left parenthesis 0 comma negative 2 right parenthesis intersect at left parenthesis 1 comma negative 2 right parenthesis, forming a right triangle with the slope.

An illustration of a coordinate plane with four quadrants shows the x-axis ranging from negative 9 to 9 and the y-axis ranging from negative 11 to 11 in increments of 1. A solid upward line with arrows at both ends is plotted on the graph. The line passes through the plotted points left parenthesis 0 comma 3 right parenthesis and left parenthesis 1 comma 5 right parenthesis.
Image with alt text: An illustration of a coordinate plane with four quadrants shows the x-axis ranging from negative 9 to 9 and the y-axis ranging from negative 11 to 11 in increments of 1. A solid upward line with arrows at both ends is plotted on the graph. The line passes through the plotted points left parenthesis 0 comma 3 right parenthesis and left parenthesis 1 comma 5 right parenthesis.

A coordinate plane shows the x-axis ranging from negative 9 to 9 in increments of 1 and the y-axis ranging from 11 to negative 11 in increments of 1. A line with arrows at both ends joins two plotted points.
The coordinates of the plotted points are as follows: left parenthesis negative 2 comma 2 right parenthesis, and left parenthesis 0 comma 1 right parenthesis.
Image with alt text: A coordinate plane shows the x-axis ranging from negative 9 to 9 in increments of 1 and the y-axis ranging from 11 to negative 11 in increments of 1. A line with arrows at both ends joins two plotted points. The coordinates of the plotted points are as follows: left parenthesis negative 2 comma 2 right parenthesis, and left parenthesis 0 comma 1 right parenthesis.

A coordinate plane shows the x-axis ranging from negative 9 to 9 in increments of 1 and the y-axis ranging from 11 to negative 11 in increments of 1. A line with arrows at both ends joins two plotted points.
The coordinates of the plotted points are as follows: left parenthesis 0 comma 2 right parenthesis, and left parenthesis 1 comma negative 1 right parenthesis.

Answer 1

Bot GPT 4-o mini answered
2 hours ago

To graph the linear equation \(y = -3x + 2\), we can identify the slope and y-intercept of the line.

Y-Intercept: The y-intercept is the point where the line crosses the y-axis. From the equation \(y = -3x + 2\), we see that when \(x = 0\), \(y = 2\). So the y-intercept is the point \((0, 2)\).
Slope: The slope is \(-3\), which indicates that for every 1 unit you move to the right along the x-axis, you move down 3 units in the y-direction.

Using the information we have:

Start at the y-intercept point \((0, 2)\).
From that point, moving 1 unit to the right (to \(x = 1\)), you go down 3 units (to \(y = -1\)). This gives the point \((1, -1)\).

To complete the graph:

From \((1, -1)\), you can again move 1 unit right (to \(x = 2\)) and move down 3 (to \(y = -4\)), giving the point \((2, -4)\).
You can also extend the line in the opposite direction: starting from the y-intercept \((0, 2)\), move left to \((-1, 5)\) (going up because of the negative slope), giving the point \((-1, 5)\).

Now, you can plot the points:

\((0, 2)\)
\((1, -1)\)
\((2, -4)\)
\((-1, 5)\)

This will give you a straight line that extends infinitely in both directions with a negative slope, moving downwards from left to right.

Based on your description, the first option could be correct if the coordinates of the plotted points fulfill the equation. However, it mentions the points \((0, -2)\) and \((1, -5)\), which do not satisfy the given equation, as those points would not be accurately placed on the line described by \(y = -3x + 2\).

Therefore, the correct depiction of \(y = -3x + 2\) should have the points \((0, 2)\) and \((1, -1)\) plotted correctly on the graph.