Question

area = $\boldsymbol{square}$ square units
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Explanation:

Step1: Identify shape and vertices

The figure is a trapezoid with vertices: $V(-9, 0)$, $W(6, 0)$, $X(6, 2)$, and the intersection of the top line with the y-axis $(0,1)$. The two parallel sides (bases) are the bottom side along the x-axis and the top side from $V$ to $X$.

Step2: Calculate lengths of the two bases

Bottom base (VW): Distance between $(-9,0)$ and $(6,0)$ is $6 - (-9) = 15$ units.
Top base (VX segment): Using the two endpoints $(-9,0)$ and $(6,2)$, the length is $\sqrt{(6 - (-9))^2 + (2-0)^2} = \sqrt{15^2 + 2^2} = \sqrt{225+4} = \sqrt{229}$. Wait, no—for a trapezoid, we use the lengths of the two parallel sides, and the height is the perpendicular distance between them. The perpendicular height is the vertical distance between the x-axis (y=0) and the line at $X$ (y=2), so height $h=2$. Alternatively, use the formula for the area of a trapezoid: $\frac{1}{2} \times (\text{length of base 1} + \text{length of base 2}) \times \text{height}$, where base 1 is the bottom side, base 2 is the top side.
Wait, simpler: The figure is a trapezoid with one base along the x-axis from $x=-9$ to $x=6$ (length 15), the other base is the line from $(-9,0)$ to $(6,2)$, but actually, the perpendicular height is the vertical distance between the two parallel sides. Wait no, the two parallel sides are the horizontal side VW (y=0) and the side VX? No, VX is not horizontal. Wait, no—this is a trapezoid with two parallel sides: the side VW (y=0, length 15) and the side from the point $(-9,0)$ to $(6,2)$ is not parallel. Wait, actually, the shape is a trapezoid with bases as the segment VW (length 15) and the segment from $(0,1)$ to $(6,2)$? No, no—let's use coordinates to calculate the area using the shoelace formula.

Step2: Apply shoelace formula

List the vertices in order: $V(-9, 0)$, $W(6, 0)$, $X(6, 2)$, $(-9,0)$ (wait no, the fourth vertex is the point where the top line meets x=-9, which is V(-9,0). Wait no, the shape is a trapezoid with vertices $(-9,0)$, $(6,0)$, $(6,2)$, and the point where the line from $(6,2)$ to $(-9,0)$ connects. Wait, no, the figure is a trapezoid with two parallel sides: the side from $(-9,0)$ to $(6,0)$ (length 15) and the side from $(-9,0)$ to $(6,2)$ is not parallel. Wait, no—this is a trapezoid where the two parallel sides are the vertical side XW (length 2) and the side from V to the y-axis? No, I made a mistake. Let's use the shoelace formula correctly with all four vertices: $V(-9, 0)$, $W(6, 0)$, $X(6, 2)$, and the point where the line VX intersects the y-axis is $(0,1)$. Wait no, the four vertices are $(-9,0)$, $(6,0)$, $(6,2)$, $(-9,0)$ is a triangle? No, no—the graph shows a line from V(-9,0) going up to X(6,2), and a vertical line from X(6,2) down to W(6,0), so the shape is a trapezoid with vertices $V(-9,0)$, $W(6,0)$, $X(6,2)$, and the line VX connects back to V. So it's a trapezoid with bases VW (length 15) and X's vertical side? No, it's a trapezoid with two parallel sides: VW (horizontal, length 15) and the segment from $(-9,0)$ to $(6,2)$ is not parallel. Wait, no—this is a trapezoid where the two parallel sides are the side XW (vertical, length 2) and the side from V to the point $(0,1)$? No, I'm overcomplicating. Use the shoelace formula for the polygon with vertices $(-9, 0)$, $(6, 0)$, $(6, 2)$: no, that's a triangle. Wait no, the graph shows a line from V(-9,0) going up to X(6,2), so the shape is a trapezoid formed by the x-axis from V to W, the vertical line from W to X, and the line from X back to V. That is a trapezoid with two parallel sides? No, that's a triangle. Wait no, no—the line from V(-9,0) goes up to the y-axis at (0,1), then to X(6,2). Oh! I see, the shape is a trapezoid with vertices $(-9,0)$, $(6,0)$, $(6,2)$, $(0,1)$, $(-9,0)$. Yes, that's a trapezoid with two parallel sides: the bottom side from $(-9,0)$ to $(6,0)$ (length 15) and the top side from $(-9,0)$ to $(0,1)$ to $(6,2)$? No, the top side is from $(-9,0)$ to $(6,2)$, which is a single line. Wait, no—the graph shows a line starting at V(-9,0), going right and up, passing through (0,1), ending at X(6,2), then a vertical line down to W(6,0), then left along the x-axis back to V(-9,0). So this is a trapezoid with bases: the base along the x-axis (length 15) and the line segment VX (length $\sqrt{(6+9)^2 + (2-0)^2} = \sqrt{225+4} = \sqrt{229}$), and the height is the perpendicular distance between these two lines.
The equation of line VX: slope $m = \frac{2-0}{6 - (-9)} = \frac{2}{15}$. Equation: $y = \frac{2}{15}(x + 9)$.
The distance from the x-axis (y=0) to this line is the height. The formula for distance from point $(x_0,y_0)$ to line $ax+by+c=0$ is $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$. Rewrite line VX: $2x -15y +18 =0$. Distance from (0,0) to this line is $\frac{|0 -0 +18|}{\sqrt{2^2 + (-15)^2}} = \frac{18}{\sqrt{229}}$.
Then area is $\frac{1}{2} \times (15 + \sqrt{229}) \times \frac{18}{\sqrt{229}}$. That's too complicated. Instead, use the shoelace formula for the polygon with vertices $(-9,0)$, $(6,0)$, $(6,2)$, $(-9,0)$? No, that's a triangle with area $\frac{1}{2} \times 15 \times 2 =15$. But the line from V to X passes through (0,1), so the shape is actually a trapezoid with vertices $(-9,0)$, $(6,0)$, $(6,2)$, $(0,1)$, $(-9,0)$.
Shoelace formula:
List the coordinates in order:
$(-9, 0)$, $(6, 0)$, $(6, 2)$, $(0, 1)$, $(-9, 0)$
Calculate sum of $x_i y_{i+1}$:
$(-9 \times 0) + (6 \times 2) + (6 \times 1) + (0 \times 0) = 0 +12 +6 +0=18$
Calculate sum of $y_i x_{i+1}$:
$(0 \times 6) + (0 \times 6) + (2 \times 0) + (1 \times -9) =0 +0 +0 -9=-9$
Area is $\frac{1}{2} |18 - (-9)| = \frac{1}{2} |27| =13.5$? No, that's not right. Wait no, the correct vertices are $(-9,0)$, $(6,0)$, $(6,2)$, and the line from $(6,2)$ to $(-9,0)$ is a straight line, so it's a triangle, area $\frac{1}{2} \times 15 \times 2=15$. But the line passes through (0,1), so the shape is a trapezoid? No, no—the graph shows a line from V(-9,0) going up to X(6,2), and a vertical line from X to W(6,0), and a horizontal line from W to V(-9,0). That is a triangle, not a trapezoid. Wait, but the line from V to X is a straight line, so the shape is a triangle with base 15 and height 2, area 15. But wait, the y-intercept is (0,1), so the area can also be calculated as the area of the trapezoid from x=-9 to x=0 (bases 0 and 1, height 9) plus the area of the trapezoid from x=0 to x=6 (bases 1 and 2, height 6).

Step2: Calculate area as sum of two trapezoids

Left trapezoid (x=-9 to x=0):
Bases: $y=0$ and $y=1$, width (height) $0 - (-9)=9$
Area: $\frac{1}{2} \times (0+1) \times9 = 4.5$
Right trapezoid (x=0 to x=6):
Bases: $y=1$ and $y=2$, width $6-0=6$
Area: $\frac{1}{2} \times (1+2) \times6 = 9$

Step3: Total area

Add the two areas: $4.5 +9=13.5$
Wait, but this contradicts the triangle area. The mistake is: the shape is not a triangle, because the line from V to X is a straight line, so the shape bounded by V, W, X is a triangle, but the graph shows a shaded area under the line from V to X, above the x-axis. Oh! Yes, the shaded area is the region under the line from V(-9,0) to X(6,2), above the x-axis. That is a trapezoid (or a triangle? No, it's a trapezoid if we consider the two parallel sides as the x-axis and the line VX, but no—wait, no, it's a trapezoid with one base being the segment from (-9,0) to (6,0) (length 15) and the other base being the segment from (-9,0) to (6,2) (length $\sqrt{229}$), but the perpendicular height is the vertical distance? No, the correct way is to use the formula for the area under a line from $x=a$ to $x=b$. The line has equation $y = \frac{2}{15}(x+9)$. The area under this line from $x=-9$ to $x=6$ is the integral of $y dx$ from -9 to 6.

Step2: Set up integral for area

$$\int_{-9}^{6} \frac{2}{15}(x+9) dx$$

Step3: Compute the integral

First, expand the integrand: $\frac{2}{15}x + \frac{18}{15} = \frac{2}{15}x + \frac{6}{5}$
Integrate term by term:
$\int \frac{2}{15}x dx = \frac{2}{15} \times \frac{x^2}{2} = \frac{x^2}{15}$
$\int \frac{6}{5} dx = \frac{6}{5}x$
Evaluate from -9 to 6:
At $x=6$: $\frac{6^2}{15} + \frac{6}{5} \times6 = \frac{36}{15} + \frac{36}{5} = \frac{12}{5} + \frac{36}{5} = \frac{48}{5}=9.6$
At $x=-9$: $\frac{(-9)^2}{15} + \frac{6}{5} \times (-9) = \frac{81}{15} - \frac{54}{5} = \frac{27}{5} - \frac{54}{5} = -\frac{27}{5}=-5.4$
Subtract: $9.6 - (-5.4)=15$
Wait, that's 15. But why did the trapezoid sum give 13.5? Because I incorrectly split the shape into two trapezoids, but actually, the line from (-9,0) to (6,2) is a straight line, so the area under it from x=-9 to x=6 is a triangle? No, no—when x=-9, y=0; x=6, y=2. So the shape is a triangle with base 15 (from x=-9 to x=6) and height 2, area $\frac{1}{2} \times15 \times2=15$. That makes sense. My earlier mistake was thinking the shape included the vertical line, but no—the shaded area is under the line VX, above the x-axis, which is a triangle (since it starts at (x=-9,y=0) and goes to (x=6,y=2), so the area is a triangle with base 15 and height 2.

Answer:

15 square units