To find the point-slope form of the equation for the line that Mr. Shaw has graphed, we need to use the point-slope formula for a line. The point-slope form of a line is given by the equation:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is a point on the line and [tex]\(m\)[/tex] is the slope of the line.
Given the function [tex]\( f(x) = -5x + 2 \)[/tex]:
1. Identify the slope [tex]\(m\)[/tex]: The slope [tex]\(m\)[/tex] is the coefficient of [tex]\(x\)[/tex] in the linear equation. Here, the function is in the form [tex]\( y = -5x + 2 \)[/tex], so the slope [tex]\( m = -5 \)[/tex].
2. Identify a point [tex]\((x_1, y_1)\)[/tex] on the line: We are given that the point [tex]\((-2, 12)\)[/tex] lies on the line.
Next, we substitute the slope [tex]\(m = -5\)[/tex] and the point [tex]\((x_1, y_1) = (-2, 12)\)[/tex] into the point-slope form equation:
[tex]\[ y - 12 = -5(x - (-2)) \][/tex]
Simplify the expression inside the parentheses:
[tex]\[ y - 12 = -5(x + 2) \][/tex]
Therefore, the point-slope form of the equation for the line Mr. Shaw has graphed is:
[tex]\[ y - 12 = -5(x + 2) \][/tex]
The correct option is:
[tex]\[ \boxed{y - 12 = -5(x + 2)} \][/tex]
