Find two quadratic functions, one that opens upward and one that opens downward, whose graphs have the given x-intercepts.

(-3, 0), (-1/2, 0)

Quadratic functions that have given x-intercepts x1 and x2 are given by the equation:

y=a(x-x1)(x-x2)

If a>0, the curve opens upwards, and if a<0, the curve opens downwards.
|a| can be any value ≠0 depending on the location of the vertex or any other point on the curve.

To find two quadratic functions with the given x-intercepts, let's call them \(f(x)\) and \(g(x)\).

For the x-intercept (-3, 0):
Since the x-intercept is (-3, 0), it is a point on the graph of the quadratic function. We can use this point to find the equation for the upward-opening quadratic function \(f(x)\).
Using the factored form of a quadratic function, the equation for \(f(x)\) will be of the form:

\[f(x) = a(x - p)(x - q)\]

where \(p\) and \(q\) are the x-intercepts.
Substituting the values into the equation:

\[f(x) = a(x - (-3))(x - q)\]
\[f(x) = a(x + 3)(x - q)\]

Now we use the second x-intercept (-1/2, 0):
Since the x-intercept is (-1/2, 0), we can again use this point to find the equation for the downward-opening quadratic function \(g(x)\).
Using the factored form, the equation for \(g(x)\) will be:

\[g(x) = b(x - r)(x - s)\]

where \(r\) and \(s\) are the x-intercepts.
Substituting the values into the equation:

\[g(x) = b(x - (-1/2))(x - s)\]
\[g(x) = b(x + 1/2)(x - s)\]

So, the two quadratic functions are:

\[f(x) = a(x + 3)(x - q)\] (upward-opening quadratic function)
\[g(x) = b(x + 1/2)(x - s)\] (downward-opening quadratic function)

Note that we still need to determine the values of \(a\), \(q\), \(b\), and \(s\) to form specific quadratic functions.

To find the two quadratic functions, one that opens upward and one that opens downward, we need to use the x-intercepts given.

For a quadratic function, the standard form is y = ax^2 + bx + c.

Let's start with the quadratic function that opens upward:
To find the quadratic function that opens upward and passes through the x-intercepts (-3, 0) and (-1/2, 0), we can use the fact that x-intercepts occur when y = 0.

Let's substitute the x-intercepts (-3, 0) and (-1/2, 0) into the equation y = ax^2 + bx + c and solve for a, b, and c:

When x = -3:
0 = a(-3)^2 + b(-3) + c
0 = 9a - 3b + c

When x = -1/2:
0 = a(-1/2)^2 + b(-1/2) + c
0 = (1/4)a - (1/2)b + c

Now we have a system of two equations:
1) 0 = 9a - 3b + c
2) 0 = (1/4)a - (1/2)b + c

We can solve this system of equations to find the values of a, b, and c. By substituting the second equation into the first equation, we can eliminate c:

0 = 9a - 3b + [(1/4)a - (1/2)b + c]
0 = (37/4)a - (11/2)b

Now we have one equation in terms of a and b:
(37/4)a - (11/2)b = 0

Simplifying this equation, we have:
37a - 22b = 0

One possible solution for a and b is a = 2 and b = 3. By substituting these values back into the second equation, we can solve for c:

0 = (1/4) * 2 - (1/2) * 3 + c
0 = 1/2 - 3/2 + c
0 = -2/2 + c
0 = -1 + c

So, c = 1.

Therefore, the quadratic function that opens upward and passes through the x-intercepts (-3, 0) and (-1/2, 0) is y = 2x^2 + 3x + 1.

Now, let's find the quadratic function that opens downward:
Since the x-intercepts are symmetrical around the y-axis, the quadratic function that opens downward will have the same x-intercepts but in reverse order.

Therefore, the quadratic function that opens downward and passes through the x-intercepts (-3, 0) and (-1/2, 0) is y = -2x^2 - 3x - 1.

So, the quadratic function that opens upward is y = 2x^2 + 3x + 1, and the quadratic function that opens downward is y = -2x^2 - 3x - 1.