Let's examine I, II, and III separately. 

Because the parabola points downward, the value of a must be less than zero. Thus, a < 0 must be true. 

Next, let's examine whether or not c < 0. The value of c is related to the y-intercept of f(x). If we let x = 0, then f(x) = f(0) = a(0) + b(0) + c = c. Thus, c is the value of the y-intercept of f(x). As we can see from the graph of f(x), the y-intercept is greater than 0. Therefore, c > 0. It is not possible for c < 0. This means choice II is incorrect. 

Lastly, we need to examine b² – 4ac, which is known as the discriminant of a quadratic equation. According to the quadratic formula, the roots of a quadratic equation are equal to the following: 

Notice, that in order for the values of x to be real, the value of b² – 4ac, which is under the square-root sign, must be greater than or equal to zero. If b² – 4ac is negative, then we are forced to take the square root of a negative number, which produces an imaginary (nonreal) result. Thus, it cannot be true that b² – 4ac < 0, and choice III cannot be correct.

Only choice I is correct.

The answer is I only. 

We are seeking the value of  when the graph of  - a parabola - reaches its vertex.

To find this value, we first find the value of . For a parabola of the equation

,

the  value of the vertex is

.

Substitute :

The height of the baseball after 1.5625 seconds will be 

 feet.

Because we are given the vertex of the parabola, the easiest way to solve this problem will involve the use of the formula of a parabola in vertex form. The vertex form of a parabola is given by the following equation:

f(x) = a(x – h)² + k, where (h, k) is the location of the vertex, and a is a constant.

Since the parabola has its vertex as (–3, 4), its equation in vertex form must be as follows:

f(x) = a(x – (–3)² + 4 = a(x + 3)² + 4

In order to complete the equation for the parabola, we must find the value of a. We can use the point (–1, –4), through which the parabola passes, in order to determine the value of a. We can substitute –1 in for x and –4 in for f(x).

f(x) = a(x + 3)² + 4

–4 = a(–1 + 3)² + 4

–4 = a(2)² + 4

–4 = 4a + 4

Subtract 4 from both sides.

–8 = 4a

Divide both sides by 4.

a = –2

This means that the final vertex form of the parabola is equal to f(x) = –2(x + 3)² + 4. However, since the answer choices are given in standard form, not vertex form, we must expand our equation for f(x) and write it in standard form.

f(x) = –2(x + 3)² + 4

= –2(x + 3)(x + 3) + 4

We can use the FOIL method to evaluate (x + 3)(x + 3).

= –2(x² + 3x + 3x + 9) + 4

= –2(x² + 6x + 9) + 4

= –2x² – 12x – 18 + 4

= –2x² – 12x – 14

The answer is f(x) = –2x² – 12x – 14.

Because the graph of f(x) = x² is symmetric about the y-axis, when we shift it downward, the points where it intersects the x-axis will be the same distance from the origin. In other words, we could say that one intercept will be (-a,0) and the other would be (a,0). The distance between these two points has to be 8, so that means that 2a = 8, and a = 4. This means that when f(x) is shifted downward, its new roots will be at (-4,0) and (4,0).

Let g(x) be the graph after f(x) has been shifted downward. We know that g(x) must have the roots (-4,0) and (4,0). We could thus write the equation of g(x) as (x-(-4))(x-4) = (x+4)(x-4) = x² - 16.

We can now compare f(x) and g(x), and we see that g(x) could be obtained if f(x) were shifted down by 16 units; therefore, the answer is 16.