a. The original square has side length x, so its area is A1 = x^2. Darnell doubles one dimension and adds 3 centimeters, so the dimensions of the new rectangle are 2x and x + 3. Thus, the area of the new rectangle is:
A2 = (2x)(x + 3) = 2x^2 + 6x
b. To graph the area equation A2 = 2x^2 + 6x, we can plot points for different values of x and connect them with a smooth curve. For example:
- When x = 0, A2 = 0.
- When x = 1, A2 = 8.
- When x = 2, A2 = 20.
- When x = 3, A2 = 36.
We can also use algebra to find the vertex of the parabola. The x-coordinate of the vertex is given by:
x = -b/2a
where a = 2 and b = 6. Thus, the x-coordinate of the vertex is x = -6/4 = -3/2. Plugging this value into the area equation, we get:
A2 = 2(-3/2)^2 + 6(-3/2) = -9
So the vertex is at (-3/2, -9), and the graph is a downward-opening parabola.
c. To find the x-intercepts of the graph, we need to find the values of x that make A2 = 0. We can do this by setting the area equation equal to zero and solving for x:
2x^2 + 6x = 0
2x(x + 3) = 0
x = 0 or x = -3
Thus, the x-intercepts of the graph are (0, 0) and (-3, 0). We can find the x-intercepts from the graph by looking for the points where the curve intersects the x-axis. We can find them from the equation by setting A2 = 0 and solving for x, as shown above.