Please help with my math hw

Question

The figure shows a parabola with a rectangle inscribed. The parabola is the graph of the function, y = f(x) = 8 – x2. The right vertex of the rectangle intersects the parabola at (x,y). Find a function which gives the area of the rectangle, A(x), as a function of x.

figure described in the question

choices:

a) A(x)=x(8−x^2)

b) A(x)=2x(8−x^2)

b) A(x)=x(16−x^2)

d) A(x)=16x(1−x^2)

Answer 1

To find the area of the rectangle, we need to determine the dimensions of the rectangle and then multiply the length and width. The rectangle is inscribed in the parabola, so it will have its sides parallel to the coordinate axes.

Let's denote the length of the rectangle as \(L(x)\) and the width as \(W(x)\). The right vertex of the rectangle intersects the parabola at \((x, y = f(x) = 8 - x^2)\).

Since the sides of the rectangle are parallel to the coordinate axes, we have:

\(L(x) = 2x\) (length along the x-axis)

\(W(x) = f(x) = 8 - x^2\) (width along the y-axis)

Now, the area of the rectangle \(A(x)\) as a function of \(x\) is given by:

\(A(x) = L(x) \cdot W(x)\)

\(A(x) = 2x \cdot (8 - x^2)\)

Thus, the function that gives the area of the rectangle \(A(x)\) as a function of \(x\) is \(A(x) = 2x \cdot (8 - x^2)\).

Answer 2

I'm assuming the inscribed rectangle goes from the intersection point with the parabola down to the x-axis like this:

So the width of the rectangle is 2x because the parabola is symmetrical ("x" to the right and "x" to the left of the line of symmetry, the y-axis). The height of the rectangle is the function value at any point "x" which is "8 - x²". And the area is the width times the height.