# Bounded Regions Assessment

## Figure 1

The first shape I have chosen is a rectangle, the formula for area is; bh=A. The rectangles base is 4 units long and the height is 6 units high. If you multiply this you get 24 units squared, this is the area. The domain of this figure is -3 is less than or equal to x which is less than or equal to 1. The integral notation of this figure is, (-3,5.) The range of this figure is -2 is less than or equal to y which is less than or equal to 4. The integral notation of this figure is, (-2, 4.) The left vertical line had the slope intercept equation of y=-3 and a point slope equation that was y-2=6/0(x+3.) The right vertical line had a slope intercept equation of y=1. The point slope form was y-0=6/0(x-1.) The top horizontal line had a slope intercept form of y=4. The point slope form was y-4=0(x-0.) The bottom horizontal line had a slope intercept equation of y=-2. The point slope form was y+2=0(x+1.)

## Figure 5

The second shape I have selected is a triangle, its area formula is A=1/2bh. The base is 8 units long and the height is 4 units high. If you multiply these you get 32 units, now we have to divide it in half. 32/2 is 16, the triangle has an area of 16 units squared. The domain of this figure is -3 is less than or equal to x which is less than or equal to 5. The integral notation for this figure is, (-3,5.) The range is 0 is less than or equal to y which is less than or equal to 4. The integral notation for this figure is, (0,4.) The diagonal line on the right had a slope intercept of y=-3/4x+5. Its point slope form was y-2=-3/4(x-4.) The left diagonal line had a slope intercept of y=2/3x+2. The point slope form was y-4=2/3(x-3.) The baseline had a slope intercept of y=0. It had a point slope form of y-0=0(x-1.)

## Figure 8

The last figure I selected is a pentagon, to find its area I divided it into two shapes, a rectangle and a triangle. The first formula I have to use is A=bh and the second is, A=1/2bh. The rectangle portion of the shape has a base of 6 and a height of 4, this multiplied is 24 units. The triangle portion has a base of 4 and a height of 2. This multiplied is 8, but after dividing in half it is 4. Now I just add the two areas together and get 28 units squared. The domain of this figure is -4 is less than or equal to x which is less than or equal to 4. The integral notation is, (-4,4.) The range is 0 is less than or equal to y which is less than or equal to 4. The integral notation is, (0,4.) The top horizontal line had a slope intercept equation of y=4 and a point slope form of y-4=0(x+1.) The bottom horizontal line had a slope intercept of y=0 and a point slope form of y-0=0(x-1.) The vertical line had a slope intercept of y=-4 and a point slope form of y-2=4/0(x+4.) The negative line at the right had a slope intercept of y=-x+6 and a point slope form of y-3=-1(x-3.) The positive line at the right had a slope intercept of y=x-2 and had a point slope form of y-1=1(x-3.)

## Critical Writing

In going through the processes for figures 1 and 5 I noticed that they were simple shapes and had simple area formulas. However with figure 8 I had to divide it into two shapes and solve those to find the total area. The shapes that I created from figure 8 were the same as figures 1 and 2, (a rectangle and triangle.) Figures 7-9 are more complicated from figures 1-6 because you have to make several equations to find the area.