SOLUTION: The area of the region under the curve of a function f(x) = ax + b on the interval [0, 4] is 16 square units. (a,b) ≠ (?,?) a. (-2,8) b. (1,2) c. (3,-2) d. (5,-6) e. (7

Algebra ->  Surface-area -> SOLUTION: The area of the region under the curve of a function f(x) = ax + b on the interval [0, 4] is 16 square units. (a,b) ≠ (?,?) a. (-2,8) b. (1,2) c. (3,-2) d. (5,-6) e. (7      Log On


   



Question 1180072: The area of the region under the curve of a function f(x) = ax + b on the interval [0, 4] is 16 square units.
(a,b) ≠ (?,?)
a. (-2,8)
b. (1,2)
c. (3,-2)
d. (5,-6)
e. (7,-9)

Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


f(x) is a linear function; the area under the curve on the interval [0,4] is a trapezoid with height 4 and bases of lengths f(0) and f(4).

The area of a trapezoid is the height times the average of the bases.

a. f(x)=-2x+8; f(0)=8, f(4)=0; area = 4(8/2) = 16

b. f(x)=x+2; f(0)=2, f(4)=6; area = 4(8/2) = 16

c. f(x)=3x-2; f(0)=-2, f(4)=10; area = 4(8/2) = 16

d. f(x)=5x-6; f(0)=-6, f(4)=14; area = 4(8/2) = 16

e. f(x)=7x-9; f(0)=-9, f(4)=19; area = 4(10/2) = 20

ANSWER: The area is NOT 16 for choice e.

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Alternative solution....

We want the area to be 16; and we know the height is 4. That means we need the sum of f(0) and f(4) to be 8.

f(0)+f(4) = (0a+b)+(4a+b) = 4a+2b

a. (a,b)=(-2,8); 4a+2b=-8+16=8
b. (a,b)=(1,2); 4a+2b=4+4=8
c. (a,b)=(3,-2); 4a+2b=12-4=8
d. (a,b)=(5,-6); 4a+2b=20-12=8
e. (a,b)=(7,-9); 4a+2b=28-18=10