Question

a function y = f(x) consists of 3 line segments as shown in the standard (x,y) coordinate plane below, where the endpoints of each segment are labelled on the graph. (2,7) (4,7) (0,2) (6,0) note: figure not drawn to scale! what is the area, in square coordinate units, of the region bounded by the graph of y = f(x), the positive x - axis, and the positive y - axis? 21 24 27 30 33

Explanation:

Step1: Calculate area of first triangle

The first - segment is a triangle with base $b_1 = 2$ and height $h_1=7 - 2=5$. Using the area formula for a triangle $A=\frac{1}{2}bh$, we have $A_1=\frac{1}{2}\times2\times5 = 5$.

Step2: Calculate area of rectangle

The second - segment is a rectangle with length $l = 4 - 2=2$ and width $w = 7$. Using the area formula for a rectangle $A=lw$, we get $A_2=2\times7 = 14$.

Step3: Calculate area of second triangle

The third - segment is a triangle with base $b_2=6 - 4 = 2$ and height $h_2 = 7$. Using the area formula for a triangle $A=\frac{1}{2}bh$, we obtain $A_3=\frac{1}{2}\times2\times7=7$.

Step4: Calculate total area

The total area $A = A_1+A_2+A_3$. Substituting the values of $A_1$, $A_2$, and $A_3$, we have $A=5 + 14+7=26$. But we made a mistake above. Let's calculate it in another way.
The region can be split into a trapezoid and a triangle.
The trapezoid has bases $b_1 = 2$ and $b_2=7$ and height $h_1 = 2$. Using the trapezoid - area formula $A_{trap}=\frac{(b_1 + b_2)h}{2}=\frac{(2 + 7)\times2}{2}=9$.
The second part is a trapezoid with bases $b_1 = 7$ and $b_2 = 0$ and height $h_2=2$, and its area $A_{trap2}=\frac{(7 + 0)\times2}{2}=7$.
The third part is a rectangle with length $l = 2$ and height $h = 7$, and its area $A_{rect}=2\times7 = 14$.
The total area $A=\frac{(2 + 7)\times2}{2}+ \frac{(7+0)\times2}{2}+2\times7$.
$A = 9+7 + 14=30$.

Answer:

30