I got 2 more intense problems like this. Can someone show me the way for each step so I can follow along with the other problems I have to do? ThankS!

A gardener has 40 feet of fencing with which to enclose a garden adjacent to a long existing wall. The gardener will use the wall for one side and the available fencing for the remaining three sides.

If the sides perpendicular to the wall have length x feet, which of the following (A, B, C, or D) represents the area A of the garden?

A. A(x) = –2x^2 + 20x
B. A(x) = –2x^2 + 40x
C. A(x) = 2x^2 – 40x
D. A(x) = x^2 – 40x

The area function is a quadratic function and so its graph is a parabola.

Does the parabola open up or down?

Find the vertex of the quadratic function A(x). Show work.

Use the work in the previous parts to help determine the dimensions of the garden which yield the maximum area, and state the maximum area. (Fill in the blanks below. Include the units of measurement.)

The maximum area is ?

when the sides perpendicular to the wall have length x = ?
and the side parallel to the wall has length ?.

http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut34_quadfun.htm

For the perpendicular sides, the distance is x; there are two sides so that is 2x if added together. Since the fence is the third side, then the fourth must be 40-2x for its length. Therefore, the area is x(40-2x).

Use the site I posted above to find the vertex and if the parabola opens up or down.

To find the answer for the first problem, which choice represents the area A of the garden, we need to determine the correct quadratic function.

Since the gardener will use the existing wall for one side, the length of the fencing used will be an L-shape. Let's denote the length of the sides perpendicular to the wall as x, and the side parallel to the wall as y.

The total length of fencing used is given as 40 feet, so we can write the equation as:

x + x + y = 40

Simplifying this equation, we find:

2x + y = 40

Now, we need to express y in terms of x so that we have the area function A(x) in terms of x only. Since the area of a rectangle is given by length times width, the width (side parallel to the wall) will be y and the length (sides perpendicular to the wall) will be x. Therefore, the area A will be equal to x * y.

To express y in terms of x, we can isolate y in the equation 2x + y = 40:

y = 40 - 2x

Now we substitute y in the expression for the area:

A(x) = x * (40 - 2x)

Simplifying further:

A(x) = 40x - 2x^2

Comparing this expression to the answer choices, we see that the correct representation of the area A is:

A(x) = -2x^2 + 40x

So, the answer is B.

Now, let's move on to the second problem, which asks whether the parabola opens up or down.

In general, a parabola opens up if the coefficient of x^2 (the quadratic term) is positive, and it opens down if the coefficient is negative.

In this case, the coefficient of x^2 is -2. Since it is negative, the parabola opens down.

Next, we need to find the vertex of the quadratic function A(x). The vertex is the point on the parabola where it reaches its maximum or minimum value.

The formula for finding the x-coordinate of the vertex of a quadratic function in the form A(x) = ax^2 + bx + c is:

x = -b/2a

In this case, the quadratic function is A(x) = -2x^2 + 40x. Therefore, the coefficient of x^2 is a = -2 and the coefficient of x is b = 40.

Substituting these values into the formula, we have:

x = -(40) / (2 * -2)
x = -40 / -4
x = 10

To find the y-coordinate of the vertex, we substitute this value of x back into the quadratic function:

A(10) = -2(10)^2 + 40(10)
A(10) = -200 + 400
A(10) = 200

Therefore, the vertex of the quadratic function A(x) is (10, 200).

Now, using the work from the previous parts, we can determine the dimensions of the garden that yield the maximum area.

We found that the length of the sides perpendicular to the wall is x = 10 feet. The length of the side parallel to the wall, y, can be found by substituting x = 10 into the equation y = 40 - 2x:

y = 40 - 2(10)
y = 40 - 20
y = 20

So, the length of the side parallel to the wall is y = 20 feet.

Therefore, the dimensions of the garden that yield the maximum area are:
- The sides perpendicular to the wall have a length of x = 10 feet
- The side parallel to the wall has a length of y = 20 feet

Finally, the maximum area can be found by substituting these values back into the expression for the area A(x):

A(10) = -2(10)^2 + 40(10)
A(10) = -200 + 400
A(10) = 200

Therefore, the maximum area is 200 square feet.