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qlt1 task 5

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qlt1 task 5
Task 2
Use the situation above to complete parts A1 through A5.
1. Find the x-intercept and y-intercept of the given equation algebraically, showing all work.
Y = -2/3x + 30 x-intercept 0=-2/3x+30 -30=-2/3X (3/2)(-30)=(-2/3x)(3/2) -45=-x 45=x x-intercept: (45,0) y-intercept y=-2/3(0)+30 y=30 y-intercept: (0,30) 2. Graph the given equation.
• Label each axis of the coordinate plane with descriptive labels.
• Label each intercept as “x-intercept” or “y-intercept” and include the ordered pair.

3. Identify the points on the graph that most accurately represent the following:
• The location of the third-story window as an ordered pair. (0,30)
• The location where the laser beam hits the ground as an ordered pair. (45,0)

4. Determine the height of the laser beam 30 feet away from the face of the building.
Y=(-2/3)(30)+30
Y=10
Height of the laser beam 30 feet away is 10 feet.

a. Explain the process used to solve this problem algebraically or graphically, showing all work. I substituted the value of 30 for “x” and solved the problem.

5. Determine which quadrant(s) is(are) relevant to this problem.
a. Explain whether the graph is a reasonable visual representation of the path of the laser beam, based on attributes of the story problem and the nature of the graph.
Quadrant #1 is relevant to this problem. The graph is a reasonable representation of this problem. The slope is a negative slope. Therefore, it is facing in the proper direction represented graphically. The man is in a 3 story building shining a laser down to the pavement. Since the man is 3 stories above ground level he must be on the positive side of the y-axis. Since he is shining the laser to the level of the pavement, the assumption is that value of height is 0 when it meets the ground. Therefore, the “y” value would be 0 at that point.

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