c (1.61,0), (1.61,3.22), (- 1.61,3.22), (- 1.61,0)
D
d 10.37
Practice makes perfect
a To find the area of the rectangle, we need to know its dimensions — length l and width w. Right away we can determine the width of the rectangle by looking at the diagram.
The width of the rectangle is 2. Now, we will find the length of the rectangle. The length is equal to the y-coordinate of the upper vertices. To find it, we will substitute x= 1 into the equation of the function.
b We will follow the same process as in Part A. By looking at the diagram, we can determine the length l of the rectangle.
The length of the rectangle is 6. Now we will find the width w of the rectangle. The width is equal to the y-coordinate of the upper vertices. To find it, we will substitute x= 3 into the equation of the function.
c We are asked to find the coordinates of the vertices of the square constructed in the same manner as the rectangles.
First, we will consider only the highlighted vertex. Let y be the side length of the square. It is the y-coordinate of the upper vertices as well. Since the y-axis cuts the square in half we receive the following equation.
y/2=x ⇔ y=2x
The vertex must be also included in the parabola. Using this information we can write the second equation.
y=- 0.3x^2+4
Let's write our system of equations.
y=2x & (I) y=- 0.3x^2+4 & (II)
We will solve this system using the Substitution Method. Note that the vertex which we are considering lays in the first quadrant. Therefore, we will only consider the coordinate pairs (x,y) where x>0 and y>0 as our solutions.
Let's solve the second equation using the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
In our case a=- 0.3, b=- 2, and c=4. Let's substitute these values into the formula.
Let's split the result into the positive and negative cases.
x=2± 2.966.../- 0.6
x=2+2.996.../- 0.6
x=2-2.966.../- 0.6
x=- 8.277...
x=1.610...
x≈ - 8.28
x≈ 1.61
Recall that we only consider the positive solutions. Therefore, our only solution is x≈ 1.61. Using the first equation from the system, y=2x, we can find the value of y.
The highlighted vertex of the square is the point (1.61,3.22). This information allows us to determine the coordinates of the remaining three vertices.
The vertices of the square are (1.61,0), (1.61,3.22), (- 1.61,3.22), and (- 1.61,0).
d We want to determine the area of the square from Part C. Let's recall the formula for the area of a square.
A=s^2
Here, s is the side length of the square. In our case s= 3.22. Let's substitute this value into the formula and simplify!
