Find two quadratic functions whose graph have the given x-intercepts. Find one function whose graph opens upward and another whose graph open downdward.

(-5/2,0),(-2,0)

upwards:

f(x) = (x+2)(2x+5)

downwards
f(x) = -(x+2)(2x+5)

expand if needed

To find two quadratic functions that have the given x-intercepts, we can start by using the fact that the x-intercepts of a quadratic function occur when the function equals zero.

Let's assume the first quadratic function opens upward. We can use the form of a quadratic function, which is generally written as follows:
f(x) = ax^2 + bx + c,

where a, b, and c are constants.

Since the x-intercepts are (-5/2, 0) and (-2, 0), we can set up equations based on these x-intercepts:

For the first x-intercept:
When x = -5/2, the function equals zero.
Substituting these values into the equation, we get:
0 = a(-5/2)^2 + b(-5/2) + c.

For the second x-intercept:
When x = -2, the function equals zero.
Substituting these values into the equation, we get:
0 = a(-2)^2 + b(-2) + c.

Now, we have two equations:
1) 0 = a(-5/2)^2 + b(-5/2) + c,
2) 0 = a(-2)^2 + b(-2) + c.

To determine the constants a, b, and c, we need one more equation, which can be obtained by specifying that the graph of the quadratic opens upward.

If a quadratic function opens upward, its leading coefficient (the coefficient of the x^2 term, which is 'a' in our case) should be positive.

Therefore, we can assume a positive value for 'a', such as 1.

Using a = 1, we can substitute this value into the above two equations to find 'b' and 'c'.

1) 0 = (1)(-5/2)^2 + b(-5/2) + c,
2) 0 = (1)(-2)^2 + b(-2) + c.

Simplifying these equations further:

1) 0 = 25/4 - 5/2b + c,
2) 0 = 4 - 2b + c.

To find the values of 'b' and 'c', we can solve this system of equations. Subtracting equation 2 from equation 1, we have:

0 = (25/4 - 4) - (5/2b - (-2b)) + (c - c),
0 = 25/4 - 16/4 - (5/2b + 2b) + 0.

Simplifying further, we get:

0 = 9/4 - (9/2b) + 0.

To isolate 'b', multiply both sides of the equation by 4:
0 = 36 - 18b.

Dividing by -18 to solve for 'b':
b = 36 / 18,
b = 2.

Now that we have the value of 'b', we can substitute it back into either equation 1 or 2 to find 'c'. Let's use equation 2:

0 = 4 - 2b + c,
0 = 4 - 2(2) + c,
0 = 4 - 4 + c,
0 = c.

Therefore, 'c' is 0.

We have found the values of 'a', 'b', and 'c':

a = 1,
b = 2,
c = 0.

So, the quadratic function that opens upward and has x-intercepts (-5/2, 0) and (-2, 0) is:
f(x) = x^2 + 2x + 0, which simplifies to f(x) = x^2 + 2x.

Now, let's find another quadratic function that opens downward with the same x-intercepts.

For a quadratic function that opens downward, the leading coefficient (the coefficient of the x^2 term, which is 'a' in our case) should be negative.

Let's assume a negative value for a, such as -1.

Using a = -1, we can substitute this value into the same two equations we obtained earlier, and solve for 'b' and 'c'.

1) 0 = (-1)(-5/2)^2 + b(-5/2) + c,
2) 0 = (-1)(-2)^2 + b(-2) + c.

Simplifying these equations further:

1) 0 = 25/4 + 5/2b + c,
2) 0 = 4 + 2b + c.

Again, subtract equation 2 from equation 1:

0 = (25/4 - 4) - (5/2b - (-2b)) + (c - c),
0 = 25/4 - 16/4 - (5/2b + 2b) + 0.

Simplifying further, we get:

0 = 9/4 - (9/2b) + 0.

To isolate 'b', multiply both sides of the equation by 4:
0 = 36 - 18b.

Dividing by -18 to solve for 'b':
b = 36 / 18,
b = 2.

Now, substitute 'b' back into equation 2:

0 = 4 + 2(2) + c,
0 = 4 + 4 + c,
0 = 8 + c.

Therefore, 'c' is -8.

The quadratic function that opens downward and has x-intercepts (-5/2, 0) and (-2, 0) is:
f(x) = -x^2 + 2x - 8.

To summarize:

The quadratic function that opens upward and has x-intercepts (-5/2, 0) and (-2, 0) is:
f(x) = x^2 + 2x.

The quadratic function that opens downward and has x-intercepts (-5/2, 0) and (-2, 0) is:
f(x) = -x^2 + 2x - 8.