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course Mth 271
10/4 830p
Sketch and completely label a trapezoidal approximation graph for the function y = x 2/ 4 + 1, for x = 0 to 1.8 by increments of .6.The given information creates a graph with points (.6, 1.09), (1.2, 1.36) and (1.8, 1.81). Using the y-intercept of (0,1) creates a 4th point.
Drawing vertical lines from the y-coordinates down to the x-axis creates 3 trapezoids.
Areas for
trapezoid 1: coordinates(0,0) (0,1)(.6, 1.09)(0,.6) area .027, slope .5
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The last point would be (.6, 0). You similarly reversed coordinates of the last point on the other trapezoids.
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You could sketch a rectangle with height 1 and width .6 inside of that first trapezoid. That rectangle would have area 1 * .6 = .6.
So it isn't possible that the trapezoid would have area .027.
The rise from the first to second point would be .09, the run would be .6. The resulting slope is not .5.
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You need to do your calculations based on a good sketch of this graph, including these trapezoids.
And you need to explain how you calculated your areas and slopes for at least the first trapezoid. Without that information I can't tell how you are calculating those quantities, so I can't address any errors, except to say that there appear to be some.
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trapezoid 2 coordinates(0,.6)(.6, 1.09)(1.2, 1.36)(0, 1.2) area .081, slope .45
trapezoid 3 coordinates(0, 1.2)(1.2,1.36)(1.8,1.81)(0,1.8)area .135, slope .75
total area = .027+.081+.135=.243
slope change 1 (.45-.5/.6 = -.083333….
slope change 2(.75-.45)/.6 = .5
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The slopes you are using don't appear to be right, but the way you are calculating the rates of slope change is valid. If you had the right slopes you would get the right rates.
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See my notes. I can't tell how you are calculating the areas and thh slopes.
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