A quadrilateral in the coordinate plane has vertices (3, -2), (3, -4), (5, -4), and (5, -2). What is the area, in square units,
1 answer:
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See the attached picture:
Problem 1
With limits, you are looking to see what happens when x gets closer to some value. For example, as x gets closer to x = 2 (from the left and right side), then y is getting closer and closer to y = 1/2. Therefore the limiting value is 1/2
Another example: as x gets closer to x = 4 from the right hand side, the y value gets closer to y = 4. This y value is different if you approach x = 0 from the left side (y would approach y = 1/2)
Use examples like this and you'll get the results you see in "figure 1"
For any function values, you'll look for actual points on the graph. A point does not exist if there is an open circle. There is an open circle at x = 2 for instance, so that's why f(2) = UND. On the other hand, f(0) is defined and it is equal to 4 as the point (0,4) is on the function curve.
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Problem 2
This is basically an extension of problem 1. The same idea applies. See "figure 2" (in the attached images) for the answers.
With limits, you are looking to see what happens when x gets closer to some value. For example, as x gets closer to x = 2 (from the left and right side), then y is getting closer and closer to y = 1/2. Therefore the limiting value is 1/2
Another example: as x gets closer to x = 4 from the right hand side, the y value gets closer to y = 4. This y value is different if you approach x = 0 from the left side (y would approach y = 1/2)
Use examples like this and you'll get the results you see in "figure 1"
For any function values, you'll look for actual points on the graph. A point does not exist if there is an open circle. There is an open circle at x = 2 for instance, so that's why f(2) = UND. On the other hand, f(0) is defined and it is equal to 4 as the point (0,4) is on the function curve.
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Problem 2
This is basically an extension of problem 1. The same idea applies. See "figure 2" (in the attached images) for the answers.