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course Mth 163
2/3 8:59 pm
If water depths of 37.6, 31, 27.1 and 25.9 cm are observed at clock times 17.8, 26.7, 35.6 and 44.5 sec, then at what average rate does the depth change during each time interval?Sketch a graph of this data set and use a sketch to explain why the slope of this graph between 26.7 and 35.6 sec represents the average rate at which depth changes during this time interval.
If f(x) = x2, give the vertex and the three basic points of the graphs of f(x--1.75), f(x) - .85, 2 f(x) and 2 f(x--1.75) + .85. Quickly sketch each graph.
Rate of time change is 8.9 sec, rated of depth change is -6.6cm, -3.9cm, and -1.2cm respectively.
Delta rise/ delta run gives -.74157, -.43820, and -.13483. Difference between 1st and second is .30337, difference between 2nd and 3rd is .30337, average rate of change is .30337.
The graph is a smooth curve with varying slope, because this is a quadratic function the average rate of depth change per time interval is the same, .30337.
f(x - 1.75) = x^2 - 3.5x + 3.0625, vertex is (1.75, 0) other two basic points are ( .75, 1) and (2.75, 1)
Concave upward parabola with given basic points.
f(x) - .85 = x^2 - .85, vertex is (0, -.85) other two basic points are (-1,.15) and (1,.15) difference in original is graph is shifted -.85 vertically.
2 f(x) = 2x^2, vertex is (0,0) other two basic points are (-1, 2) and (1, 2) graph is stretched 2 times as steep putting y coordinates twice as far as original graph.
2 f(x - 1.75) + .85 = 2x^2 - 7x + 6.975, vertex is ( 1.75, .85) other two basic points are (.75, 2.85) and (2.75, 2.85), graph is shifted vertically .85, horizontally 1.75 and twice as steep as original.
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