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Which linear function equation would contain the points below? (-6,-8) and (12,4)Asked by Oakley
Which linear function equation would contain the points below?
(-6,-8) and (12,4)
(-6,-8) and (12,4)
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Answered by
GPT-4o mini
To find the linear function that contains the points \((-6, -8)\) and \( (12, 4)\), we can start by finding the slope of the line connecting these two points. The slope \(m\) is calculated using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Substituting the coordinates of the points \((-6, -8)\) and \( (12, 4)\):
- \( (x_1, y_1) = (-6, -8) \)
- \( (x_2, y_2) = (12, 4) \)
Now we plug these values into the slope formula:
\[
m = \frac{4 - (-8)}{12 - (-6)} = \frac{4 + 8}{12 + 6} = \frac{12}{18} = \frac{2}{3}
\]
Now that we have the slope, we can use the point-slope form of the linear equation:
\[
y - y_1 = m(x - x_1)
\]
Using one of the points, let’s use \((-6, -8)\):
\[
y - (-8) = \frac{2}{3}(x - (-6))
\]
\[
y + 8 = \frac{2}{3}(x + 6)
\]
Now we can simplify this equation:
\[
y + 8 = \frac{2}{3}x + 4
\]
\[
y = \frac{2}{3}x + 4 - 8
\]
\[
y = \frac{2}{3}x - 4
\]
Thus, the equation of the linear function that contains the points \((-6, -8)\) and \((12, 4)\) is:
\[
y = \frac{2}{3}x - 4
\]
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]
Substituting the coordinates of the points \((-6, -8)\) and \( (12, 4)\):
- \( (x_1, y_1) = (-6, -8) \)
- \( (x_2, y_2) = (12, 4) \)
Now we plug these values into the slope formula:
\[
m = \frac{4 - (-8)}{12 - (-6)} = \frac{4 + 8}{12 + 6} = \frac{12}{18} = \frac{2}{3}
\]
Now that we have the slope, we can use the point-slope form of the linear equation:
\[
y - y_1 = m(x - x_1)
\]
Using one of the points, let’s use \((-6, -8)\):
\[
y - (-8) = \frac{2}{3}(x - (-6))
\]
\[
y + 8 = \frac{2}{3}(x + 6)
\]
Now we can simplify this equation:
\[
y + 8 = \frac{2}{3}x + 4
\]
\[
y = \frac{2}{3}x + 4 - 8
\]
\[
y = \frac{2}{3}x - 4
\]
Thus, the equation of the linear function that contains the points \((-6, -8)\) and \((12, 4)\) is:
\[
y = \frac{2}{3}x - 4
\]
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