Let A denote the area enclosed by the graph $f ( x ) = 4 - | x |,$ the x-axis, and the lines x = -2 and x = 0. Graphing the region and using plane geometry, we can find that A is

A)2
B)3
C)4
D)5
E)6

Question

Let A denote the area enclosed by the graph $f ( x ) = 4 - | x |,$ the x-axis, and the lines x = -2 and x = 0. Graphing the region and using plane geometry, we can find that A is

A)2
B)3
C)4
D)5
E)6

Quizplus · Accepted Answer

The function $f(x) = 4 - |x|$ is a V-shaped graph with its vertex at (0, 4). The area A is enclosed by this graph, the x-axis, and the lines x = -2 and x = 0. This forms a right triangle with a base of 2 units (from x = -2 to x = 0) and a height of 4 units (from the x-axis to the point (0, 4) on the graph). The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$, so in this case, A = $\frac{1}{2} \times 2 \times 4 = 4$. However, considering the symmetry of the graph for the region from x = 0 to x = 2, the total area enclosed by the graph, the x-axis, and the lines x = -2 and x = 2 is double that, which is $4 \times 2 = 8$. The question, however, only asks for the area from x = -2 to x = 0, which is indeed 4, indicating a mistake in my explanation. The correct interpretation should consider the entire specified region, which is indeed a single triangle with area 4, not requiring doubling. My correction thus points to an error in doubling the area for the question's scope.

Let a Denote the Area Enclosed by the Graph f(x)=4−∣x∣,f ( x ) = 4 - | x |,f(x)=4−∣x∣,

Let a Denote the Area Enclosed by the Graph $f ( x ) = 4 - | x |,$