Question

1. graph the portion of y = -0.8x + 10 that lies in the first quadrant using the slope and y - intercept.

Explanation:

Step1: Identify y - intercept

The equation is in slope - intercept form \(y = mx + b\), where \(b\) is the y - intercept. For \(y=-0.8x + 10\), when \(x = 0\), \(y=10\). So the y - intercept is the point \((0,10)\). Plot this point on the graph.

Step2: Identify slope

The slope \(m=-0.8=-\frac{4}{5}\). This means for every 5 units increase in \(x\), \(y\) decreases by 4 units. Starting from the y - intercept \((0,10)\), move 5 units to the right (increase \(x\) by 5) and 4 units down (decrease \(y\) by 4) to get the point \((5,6)\). Plot this point.

Step3: Find x - intercept

Set \(y = 0\) in the equation \(y=-0.8x + 10\). Then \(0=-0.8x + 10\), \(0.8x=10\), \(x=\frac{10}{0.8}=12.5\). So the x - intercept is the point \((12.5,0)\).

Step4: Draw the line

Draw a straight line passing through the points \((0,10)\), \((5,6)\) and \((12.5,0)\) but only keep the part of the line that lies in the first quadrant (where \(x\geq0\) and \(y\geq0\)).

Answer:

Graph a line with y - intercept at \((0,10)\), slope \(-0.8\), x - intercept at \((12.5,0)\) and keep only the part in the first quadrant.