Help for a Algebra II/Precalculus Question?

a) I would not interpret the 15 days as counted after the 10 days (giving 25), but as 15 days from the date of the first sample. That's how I read it anyway.

The data points are, with number of days in first coordinate and number of bacteria per cc in second, as follows:

(0, 900)

(10, 1600)

(15, 1875)

Note that the first sample is taken after 0 days :)

b) It is important to use both coordinates. So draw an X-axis and Y-axis, where the X-axis has steps of 5 units (days), and the Y-axis has steps of 100 units (bacteria/cc). You can draw the points approx. like this:

1900┼      ●

1800┼

1700┼

1600┼    ●

1500┼

1400┼

1300┼

1200┼

1100┼

1000┼

0900●

0800┼

0700┼

0600┼

0500┼

0400┼

0300┼

0200┼

0100┼

0000┼─┼─┼─┼─┼─┼─┼─┼ X-axis (time since first sample, expressed in days)

    0  5 10 15

c) a parabola has the form y = ax² + bx + c

Given the data points from (a), this equation should hold for the first data point:

(x,y) = (0,900)

So: a·0² + b·0 + c = 900, or

c = 900

So we have now as equation y = ax² + bx + 900. This equation should hold for the second data point:

(x,y) = (10,1600)

So: a·10² + b·10 + 900 = 1600, or

100a + 10b = 700, or

b = 70 - 10a

So we have now as equation y = ax² + (70 - 10a)x + 900. This equation should hold for the third data point:

(x,y) = (15,1875)

So: a·15² + (70-10a)·15 + 900 = 1875, or

225a + 1050 - 150a + 900 = 1875, or

75a = -75, or

a = -1

The equation of the parabola thus becomes: y = -x² + 80x + 900

d) Elimination is when y = 0 (i.e. 0 bacteria per cc). So solve this:

0 = -x² + 80x + 900

Solutions are x = [-b ± √(b²-4ac)] / 2a, or

x = [-80 ± √(80² + 4·900) ] / (-2), or

x = 40 ± √(6400 + 3600) / 2, or

x = 40 ± √(10000) / 2, or

x = 40 ± 100 / 2, or

x = 90 or x = -10

Since we are only interested in the positive result, the answer is 90 days.

e) The answer to this is the other solution found in (d): 10 days before the first sample was taken he got infected.

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