Graph the equation 9−x=32y using slope-intercept form.(1 point)

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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 negative 3 right parenthesis, and left parenthesis 7 comma negative 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 0 comma negative 3 right parenthesis, and left parenthesis 7 comma negative 1 right parenthesis.

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

An illustration shows a coordinate plane with four quadrants. The x and y axes range from negative 10 to 10 in one unit increments. A downward slanting line is graphed on the coordinate plane. The downward slanting line passes through closed points plotted at left parenthesis 0 comma 6 right parenthesis and left parenthesis 3 comma 4 right parenthesis.
Image with alt text: An illustration shows a coordinate plane with four quadrants. The x and y axes range from negative 10 to 10 in one unit increments. A downward slanting line is graphed on the coordinate plane. The downward slanting line passes through closed points plotted at left parenthesis 0 comma 6 right parenthesis and left parenthesis 3 comma 4 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 3 right parenthesis, and left parenthesis 4 comma 2 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 0 comma 3 right parenthesis, and left parenthesis 4 comma 2 right parenthesis.
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Answer 1

To graph the equation \(9 - x = 32y\) using slope-intercept form, we first need to rearrange it into the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

Starting with the original equation: \[ 9 - x = 32y \]

We can rearrange it: \[ 32y = 9 - x \]

Now, isolate \(y\): \[ y = \frac{9 - x}{32} \] \[ y = -\frac{1}{32}x + \frac{9}{32} \]

Now we have the slope-intercept form \(y = mx + b\) with:

• Slope \(m = -\frac{1}{32}\)
• y-intercept \(b = \frac{9}{32}\)

To find some points that lie on this line, we can choose a few values for \(x\):

1. When \( x = 0 \): \[ y = \frac{9}{32} \approx 0.28125 \]

   So, one point is \( \left(0, \frac{9}{32}\right) \).

2. When \( x = 32 \): \[ y = -\frac{1}{32}(32) + \frac{9}{32} = -1 + \frac{9}{32} = \frac{9 - 32}{32} = -\frac{23}{32} \approx -0.71875 \]

   So, we get another point \( (32, -\frac{23}{32}) \).

Using these points, you can sketch the line on a coordinate plane.

From the options you presented, the correct graph should show a line with a very slight negative slope, given that the slope is \(-\frac{1}{32}\), making it almost flat. The y-intercept at approximately \(0.28125\) would also indicate that the line starts just above the x-axis.

None of the provided options appear to match this description, so it seems necessary to calculate the points carefully to find an appropriate graph. If a graph does not match with these values, you should choose an answer based on how close the plotted points are in corresponding placement with the negative slope and the values traversed.