What is the area, in square units, of the triangle bounded by y=0, y=x+4 and x+3y=12 ?
Hello Guest!
The zeros are -4 and 12.
Function 2 and 3 intersect in P (0,4.)
The area of the triangel is \(A=\frac{1}{2}\cdot (4+12)\cdot 4\ square\ units\)
\(A=32\ square\ units\)
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