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6) slope - intercept form $y = mx + b$ $y - 0 = \\frac{-2}{3}(x - 6)$ $…

Question

  1. slope - intercept form

$y = mx + b$
$y - 0 = \frac{-2}{3}(x - 6)$
$y = \frac{-2}{3}x + 4$
how is the slope - intercept form derived from the point - slope form in the video?
by subtracting the slope
by multiplying the slope by the y - intercept

Explanation:

Brief Explanations

To derive slope - intercept form (\(y = mx + b\)) from point - slope form (\(y - y_1=m(x - x_1)\)), we use the distributive property (multiply the slope \(m\) by the terms inside the parentheses \((x - x_1)\)) and then solve for \(y\). For example, in the given point - slope form \(y - 0=-\frac{2}{3}(x - 6)\), we distribute \(-\frac{2}{3}\) to \(x\) and \(- 6\) (i.e., \(y=-\frac{2}{3}x+(-\frac{2}{3})\times(-6)\)), simplify the constant term (\((-\frac{2}{3})\times(-6) = 4\)) to get \(y =-\frac{2}{3}x + 4\), which is slope - intercept form. So the process involves distributing (multiplying the slope by the quantity in the parentheses) and then isolating \(y\). From the given options, the correct way is by distributing (multiplying the slope by the \(x\) - term and the constant term from the point - slope form's parentheses) and then solving for \(y\), which aligns with the idea of multiplying the slope by the terms in the point - slope form's \((x - x_1)\) part and then simplifying to get \(y=mx + b\).

Answer:

To derive slope - intercept form from point - slope form, we use the distributive property (multiply the slope \(m\) by \((x - x_1)\) in \(y - y_1=m(x - x_1)\)) and then solve for \(y\). For example, starting with \(y - 0=-\frac{2}{3}(x - 6)\), distribute \(-\frac{2}{3}\): \(y=-\frac{2}{3}x+(-\frac{2}{3})\times(-6)\), simplify \((-\frac{2}{3})\times(-6) = 4\) to get \(y =-\frac{2}{3}x + 4\). So the correct method involves multiplying the slope by the terms in the point - slope form's parentheses (distributive property) and then isolating \(y\). If we consider the given (partially visible) options, the correct approach is related to multiplying the slope by the appropriate terms (from the point - slope form) and then solving for \(y\) to get \(y=mx + b\).