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

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

(b) Determine $f(0)$.

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

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

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-034.jpg?height=485&width=439&top_left_y=1141&top_left_x=1364)
(a) The line $r$ passes through the point $(0,2)$, so its equation is given by $y=m x+2$. Since this line also passes through the point $(-2,0)$, we have $0=-2 m+2$, which implies $m=1$. Thus, the equation of $r$ is $y=x+2$. The line $s$ passes through the point $(0,6)$, so its equation is given by $y=m x+6$ and, as it also passes through the point $(3,0)$, we have $0=3 m+6$, which implies $m=-2$. Thus, the equation of $s$ is $y=-2 x+6$.
(b) Let $A$ be the intersection point of the lines $r$ and $s, B=(0,2), O=(0,0)$, $D=(3,0)$ and let $C$ be the intersection point of the line $s$ with the horizontal line through $B$, as shown in the given figure. By definition, $f(0)$ is the sum of the areas of the triangle $\triangle A B C$ and the trapezoid $B O D C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-116.jpg?height=529&width=448&top_left_y=495&top_left_x=878)

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

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

(c) Observe that $f(y)$ is equal to $f(0)$ minus the area of the trapezoid of height $y$ and bases 3 and $x$, where $x$ is the abscissa of the point on the line $s$ that has ordinate $y$, i.e., satisfies $y=-2 x+6$. Thus, $x=(6-y) / 2=3-\frac{1}{2} y$ and the area of this trapezoid is given by

$$
\frac{3+x}{2} y=\frac{3+3-\frac{1}{2} y}{2} y=3 y-\frac{1}{4} y^{2}
$$

and we obtain, for $0 \leq y<2$,

$$
f(y)=\frac{19}{3}-3 y+\frac{1}{4} y^{2}=\frac{1}{4} y^{2}-3 y+\frac{19}{3}
$$

(c) The graph of $f(y)=\frac{y^{2}}{4}-3 y+\frac{19}{3}$ is a parabola concave upwards. The coordinates of the vertex $V$ of this parabola are $x=\frac{3}{2 / 4}=6$ and

$$
y=f(6)=\frac{6^{2}}{4}-3 \times 6+\frac{19}{3}=-9+\frac{19}{3}=-\frac{8}{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-116.jpg?height=366&width=382&top_left_y=2164&top_left_x=1431)

i.e., $V=(6,-8 / 3)$. Since $f(2)=1-6+\frac{19}{3}=\frac{4}{3}$, the graph of $f$, with $0 \leq y<2$, is the bold segment of the parabola in the given figure.