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Darnell makes a rectangle from a square by doubling one

dimension and adding 3 centimeters. He leaves the other
dimension unchanged.
a. Write an equation for the area A of the new rectangle in terms of
the side length x of the original square.

b. Graph your area equation.

c. What are the x-intercepts of the graph? How can you find the
x-intercepts from the graph? How can you find them from
the equation?

Darnell makes a rectangle from a square by doubling one dimension and adding 3 centimeters-example-1
User Charles
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1 Answer

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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.
User BadZen
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