Graphing Linear Equations: math test solutions, Grade 8 – download the PDF

1 Slope and intercept / 3 pts

1. \(m = 6\) and \(b = -8\). (1 pt)
2. \(m = -\dfrac{1}{4}\) and \(b = 3\). (1 pt)
3. It rises, because the slope \(6\) is positive. (1 pt)

2 Table and graph / 4 pts

1. \(x = -2\): \(8\). \(x = -1\): \(5\). \(x = 0\): \(2\). \(x = 1\): \(-1\). \(x = 2\): \(-4\). (2 pts)

   \(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
   \(y\)	\(8\)	\(5\)	\(2\)	\(-1\)	\(-4\)

2. Plot the five points and draw the line through them with arrowheads. (2 pts)

3 Equation from a graph / 4 pts

1. \(m = \dfrac{1 - 3}{4 - 0} = -\dfrac{2}{4} = -\dfrac{1}{2}\). (2 pts)
2. \(b = 3\), so \(y = -\dfrac{1}{2}x + 3\). (1 pt)
3. \(y = -\dfrac{1}{2}(10) + 3 = -2\). (1 pt)

4 Horizontal and vertical lines / 3 pts

1. \(y = 6\). (1 pt)
2. \(x = -3\). (1 pt)
3. The horizontal line has slope \(0\); the slope of the vertical line is undefined. (1 pt)

5 Rate of change from a table / 3 pts

When \(x\) goes up by 2, \(y\) goes up by 6 each time, so the change is constant. (1 pt)

\(m = \dfrac{11 - 5}{3 - 1} = 3\). (1 pt)

From \((1, 5)\): \(5 = 3(1) + b\), so \(b = 2\). The equation is \(y = 3x + 2\). Check with \(x = 7\): \(23\). (1 pt)

6 Bike rental / 3 pts

1. \(C = 6h + 8\). (1 pt)
2. \(C = 6(5) + 8 = 38\), so $38. (1 pt)
3. \(6h + 8 = 50\), so \(6h = 42\) and \(h = 7\) hours. (1 pt)