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16.09. Draw the graphs of the line y = 2 x and the parabola given by y = ( x – 4) 2 on the same diagram. Calculate the area enclosed by the line and the parabola. y. 16. =. y. 2. x. 2. =. -. y. (. x. 4. ). x. 4. Draw the graphs. 16.09. - PowerPoint PPT Presentation
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Theta Mathematics Workbook (NCEA L2)© Pearson Education New Zealand 2005
x
y16
16.09
Draw the graphs of the line y = 2x and the parabola given by y = (x – 4)2 on the same diagram. Calculate the area enclosed by the line and the parabola.
Draw the graphs. xy 2
2)4( xy
4
Theta Mathematics Workbook (NCEA L2)© Pearson Education New Zealand 2005
x
y16
4
16.09
Draw the graphs of the line y = 2x and the parabola given by y = (x – 4)2 on the same diagram. Calculate the area enclosed by the line and the parabola.
To find the area between the line and the parabola, we first need to find the x-values of the points of intersection.
Solve the equations y = 2x and y = (x – 4)2 simultaneously.
Theta Mathematics Workbook (NCEA L2)© Pearson Education New Zealand 2005
x
y16
4
16.09
Draw the graphs of the line y = 2x and the parabola given by y = (x – 4)2 on the same diagram. Calculate the area enclosed by the line and the parabola.
To find the area between the line and the parabola, we first need to find the x-values of the points of intersection.
xx 2)4( 2
xxx 21682
016102 xx
0)8)(2( xx
8or 2 xx2 8
Theta Mathematics Workbook (NCEA L2)© Pearson Education New Zealand 2005
x
y16
4
16.09
Draw the graphs of the line y = 2x and the parabola given by y = (x – 4)2 on the same diagram. Calculate the area enclosed by the line and the parabola.
2 8
The area shadedis the area under the parabola.
The combination of bothis the area under the line.
The required area is shaded
Look at the areas shaded on the graph.
Thus the required area is the difference between the area under the line between x = 2 and x = 8 and the area under the parabola.
Theta Mathematics Workbook (NCEA L2)© Pearson Education New Zealand 2005
x
y16
4
16.09
Draw the graphs of the line y = 2x and the parabola given by y = (x – 4)2 on the same diagram. Calculate the area enclosed by the line and the parabola.
Area
2 8
3216
38
1282563
512)464(
dxx8
22
8
2
2 )168( dxxx
Under line Under parabola
3216128256
3504
60
1443
50460
14416860
8
2
23
1643
xx
x 822x
units. square36