If \( (-8,22) \) and \( (3,0) \) are two anchor points on a trend line, then find the equation of the line. \( y=[?] x+ \)
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To find the equation of the trend line through the points \( (-8, 22) \) and \( (3, 0) \), we first need to calculate the slope \( m \) using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Here, \( (x_1, y_1) = (-8, 22) \) and \( (x_2, y_2) = (3, 0) \). Calculating the slope: \[ m = \frac{0 - 22}{3 - (-8)} = \frac{-22}{3 + 8} = \frac{-22}{11} = -2 \] Now that we have the slope \( m = -2 \), we can use point-slope form \( y - y_1 = m(x - x_1) \) with one of the points, let's use \( (3, 0) \): \[ y - 0 = -2(x - 3) \] Simplifying this gives us: \[ y = -2x + 6 \] Thus, the equation can be stated in the desired format as \( y = -2x + 6 \). Therefore, the final output is: \[ y = -2x + 6 \]