If the area (in square units) of the region under the curve of the function f(x) = 4 on the interval [1, a] is 20 square units,
2 answers:
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See the attached image for the graph. Specifically figure 2 is the graph you want. You can leave the red points on the graph or decide to erase them (leave behind the blue line though).
To generate each of the red points, you'll plug in various x values to get corresponding y values.
For instance, plug in x = 0 and we get...
y = -|x-6| - 6
y = -|0-6| - 6
y = -|-6| - 6
y = -6 - 6
y = -12
So when x = 0, the y value is -12. The x and y value pair up to get (x,y) = (0,-12)
Another example: plug in x = 2
y = -|x-6| - 6
y = -|2-6| - 6
y = -|-4| - 6
y = -4 - 6
y = -10
So the point (2,-10) is on the graph
The idea is to generate as many points as possible so we get an idea of what this thing looks like.
Generate enough points, and you'll get what you see in Figure 1 (see attached image)
Then draw a line through all of the points. The more points you use, the more accurate the drawing. Doing that will generate the blue function curve you see in Figure 2 (also attached)
To generate each of the red points, you'll plug in various x values to get corresponding y values.
For instance, plug in x = 0 and we get...
y = -|x-6| - 6
y = -|0-6| - 6
y = -|-6| - 6
y = -6 - 6
y = -12
So when x = 0, the y value is -12. The x and y value pair up to get (x,y) = (0,-12)
Another example: plug in x = 2
y = -|x-6| - 6
y = -|2-6| - 6
y = -|-4| - 6
y = -4 - 6
y = -10
So the point (2,-10) is on the graph
The idea is to generate as many points as possible so we get an idea of what this thing looks like.
Generate enough points, and you'll get what you see in Figure 1 (see attached image)
Then draw a line through all of the points. The more points you use, the more accurate the drawing. Doing that will generate the blue function curve you see in Figure 2 (also attached)
In the smaller triangle, let be the length of the smallest leg. Then the smallest leg in the larger triangle is
.
Within the scope of the smaller triangle, we have such that
Then within the larger the triangle, we would have
Now we can solve for :