The function $f$ is defined for each $y, 0 \leq y<2$, such that $f(y)=$ area of the shaded quadrilateral, as indicated in the figure below.

(a) Write the equations of the lines $r$ and $s$.

(b) Determine $f(0)$.

(c) Write the expression of $f(y), 0 \leq y<2$.

(d) Sketch the graph of $f(y)$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-08.jpg?height=447&width=499&top_left_y=1625&top_left_x=1171)
(a) The line $r$ passes through the point $(0,2)$, so it has the equation $y=m x+2$. Since it also passes through the point $(-2,0)$, we verify that $0=-2 m+2$, which implies $m=1$. Thus, $r$ has the equation $y=x+2$.

The line $s$ passes through the point $(0,6)$, so $y=m x+6$ and since it also passes through the point $(3,0)$, we verify that $0=3 m+6$, which implies $m=-2$. Therefore, $s$ has the equation $y=-2 x+6$.

(b) $f(0)$ is the area of the triangle $\triangle A B C$ plus the area of the trapezoid $B O C D$, where $A$ is the intersection point of $r$ and $s$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-32.jpg?height=618&width=613&top_left_y=1187&top_left_x=672)

To determine $A$, we solve: $x+2=-2 x+6$ from which $x=4 / 3$. Substituting this value into the equation of $r$ or $s$ gives $y=10 / 3$. Thus, $A=(4 / 3,10 / 3)$. The height of the triangle $\triangle A B C$, relative to the base $B C$, is $h=10 / 3-2=4 / 3$. The point $C$ lies on the line $s$ and has $y=2$, so we have $2=-2 x+6$ or $x=2$. Therefore, $C=(2,2)$. Thus, the area of the triangle $\triangle A B C$ is $2 \times \frac{4}{3} \times \frac{1}{2}=\frac{4}{3}$ and the area of the trapezoid $B O C D$ is $2 \times \frac{3+2}{2}=5$. Therefore,

$$
f(0)=5+\frac{4}{3}=\frac{19}{3}
$$

(c) $f(y)$ is equal to $f(0)$ minus the area of the trapezoid with height $y$ and bases 3 and $x$, where $x$ is the abscissa of the point on the line $s$ that has ordinate $y$, so

$$
x=\frac{6-y}{2} \text {. }
$$

Thus, we have

$$
f(y)=\frac{19}{3}-\frac{3+\frac{6-y}{2}}{2} y=\frac{19}{3}-\frac{12 y-y^{2}}{4}=\frac{y^{2}}{4}-3 y+\frac{19}{3}
$$

(d)

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-33.jpg?height=345&width=399&top_left_y=508&top_left_x=389)

The graph of $f(y)=\frac{y^{2}}{4}-3 y+\frac{19}{3}$ is a parabola concave upwards. The coordinates of the vertex $V$ are: $x=\frac{3}{\frac{2}{4}}=6$ and $y=f(6)=\frac{6^{2}}{4}-3.6+\frac{19}{3}=-9+\frac{19}{3}=-\frac{8}{3}$. Thus, $V=\left(6,-\frac{8}{3}\right)$.

Since $f(2)=\frac{4}{3}$, the graph of $f$, for $0 \leq y \leq 2$, is the part in bold line.