AIDS Cases From 1993 to 2003 the cumulative number N of AIDS cases in thousands

Question

AIDS Cases From 1993 to 2003 the cumulative number

N of AIDS cases in thousands can be approximated by N = -2x^2 + 76x + 430, Where x = 0 corresponds to the year 1993.

Year 1993 1995 1997 1999 2001 2003

Cases 422 565 677 762 844 930



(a) Use the equation to find N for each year in the table.

(b) Discuss how well this equation approximates the data.

(c) Rewrite the equation with the right side completely

factored.

(d) Use your equation from part (c) to find N for each year

in the table. Do your answers agree with those found in

part (a)?

Explanation / Answer

(a) Use the equation to find N for each year in the table.

For 1993, putting x=0, N = -2*0^2+76*0+430 = 430

For 1995, putting x=2, N = -2*2^2+76*2+430 = 574

For 1997, putting x=4, N = -2*4^2+76*4+430 = 702

For 1999, putting x=6, N = -2*6^2+76*6+430 = 814

For 2001, putting x=8, N = -2*8^2+76*8+430 = 910

For 2003, putting x=10, N=-2*10^2+76*10+430 = 990

(b) Discuss how well this equation approximates the data.

initially very good approximation

, but for further dates from 1993, the approximation gradually decreases

becomes a little more off the actual number of cases.

(c) Rewrite the equation with the right side completely factored.

-2x^2 + 76x + 430

= -2x^2 + 86x - 10x + 430

= -2x(x-43) - 10(x-43)

= (-2x-10) (x-43)

= -2(x+5)(x-43)

(d) Use your equation from part (c) to find N for each year in the table. Do your answers agree with those found in part (a)?

For 1993, putting x=0, N=-2(0-43)(0+5)= 430

For 1995, putting x=2, N=-2(2-43)(2+5) = 574

For 1997, putting x=4, N=-2(4-43)(4+5) = 702

For 1999, putting x=6, N=-2(6-43)(6+5) = 814

For 2001, putting x=8, N=-2(8-43)(8+5) = 910

For 2003, putting x=10, N=-2(10-43)(10+5) = 990

Thus, answers agree with those found in part (a).

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