Math  /  Algebra

Question11. y=x2y=-x-2 \begin{tabular}{|c|c|c|c|} \hlinexx & x2-x-2 & yy & (x,y)(x, y) \\ \hline 0 & & & \\ \hline 1 & & & \\ \hline 2 & & & \\ \hline \end{tabular}
SLOPE /m= rise  run =/ m=\frac{\text { rise }}{\text { run }}= At what point does the line cross the yy-axis/b/y-intercept = \qquad

Studdy Solution
Identify the y-intercept of the line. The y-intercept is the constant term in the equation y=x2 y = -x - 2 .
The y-intercept is the point where x=0 x = 0 , which we calculated in STEP_1 as y=2 y = -2 .
The y-intercept is (0,2) (0, -2) .
The completed table is: xx2y(x,y)022(0,2)133(1,3)244(2,4)\begin{array}{|c|c|c|c|} \hline x & -x-2 & y & (x, y) \\ \hline 0 & -2 & -2 & (0, -2) \\ \hline 1 & -3 & -3 & (1, -3) \\ \hline 2 & -4 & -4 & (2, -4) \\ \hline \end{array}
The slope is m=1 m = -1 , and the y-intercept is (0,2) (0, -2) .

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