Graphically, the y-intercept is the point at which the graph touches the y-axis.  Algebraically, it is the value of  when .

Here, we are given the function .  In order to calculate the y-intercept, set  equal to zero and solve for .

So the y-intercept is at .

Graphically, the x-intercept is the point at which the graph touches the x-axis.  Algebraically, it is the value of  for which .

Here, we are given the function .  In order to calculate the x-intercept, set  equal to zero and solve for .

So the x-intercept is at .

A line is defined by any two points on the line.  It is frequently simplest to calculate two points by substituting zero for x and solving for y, and by substituting zero for y and solving for x.

Let .  Then

So our first set of points (which is also the y-intercept) is 

Let .  Then

So our second set of points (which is also the x-intercept) is .

Let l = length and w = width of the pen. Let us assume that the side blocked by the mountain is along the length of the pen. 

The length of wire used to make the pen must equal l + 2w, because this is the perimeter of a rectangle, excluding one of the lengths. The area of the pen will equal l x w.

l + 2w = 20

l = 20-2w

A = l x w = (20-2w)(w) = 20w - 4w²

Let A be a function of w, such that A(w) = 20w - 4w². We want to find the maximum value of A. We recognize that the graph of A must be in the shape of a parabola, pointing downward. The maximum value of the parabola will thus occur at the vertex.

We want to rewrite A(w) in the standard form of a parabola, given by f(x) = a(x-h)²+k. In order to do this, we must complete the square. 

20w-4w² = -4w²+20w = -4(w²-5w) = -4(w²-5w + 25/4) + 25 = -4(w-5/2)²+25

Thus, the vertex of the parabola occurs at (5/2, 25), which means that w = 5/2.

Going back to our original equation, l + 2(5/2) = 20, and l = 15.

A = l x w = 15(5/2) = 75/2

The function is given in vertex form, which is (x-h)²+k where the vertex of the parabola is the point (h,k).  In this particular function, h=-4 and k=3, so the vertex is (-4,-3).  No parabola is one-to-one, as they don't pass the horizontal line test.  While parabolas can be even functions, this will only happen when the vertex is on the y-axis (or when h=0) because even functions must be symmetric with respect to the y-axis.  No function can have two y-intercepts, as it would then not pass the vertical line test and not be a function.  This parabola does have two x-intercepts, however.  This can be shown by setting y=0 and solving for x, or by simply realizing that the vertex is below the x-axis and the parabola opens up.

f(x) must be a parabola, since it contains an x² term. We are told that the vertex is below the x-axis, and that the focus is below the vertex. Because a parabola always opens toward the focus, f(x) must point downward. The general graph of the parabola must have a shape similar to this: 

Since the parabola points downward, the value of a must be less than zero. Also, since the parabola points downward, it must intersect the y-axis at a point below the origin; therefore, we know that the value of the y-coordinate of the y-intercept is less than zero. To find the y-coordinate of the y-intercept of f(x), we must find the value of f(x) where x = 0. (Any graph intersects the y-axis when x = 0.) When x = 0, f(0) = a(0) + b(0) + c = c. In other words, c represents the value of the y-intercept of f(x), which we have already established must be less than zero. To summarize, a and c must both be less than zero.

The last number we must analyze is b. One way to determine whether b must be negative is to assume that b is NOT negative, and see if f(x) still has a vertex below the x-axis and a focus below the vertex. In other words, let's pretend that b = 1 (we are told b is not zero), and see what happens. Because we know that a and c are negative, let's assume that a and c are both –1.

If b = 1, and if a and c = –1, then f(x) = –x² + x – 1. 

Let's graph f(x) by trying different values of x. 

If x = 0, f(x) = –1.

If x = 1, f(x) = –1. 

Because parabolas are symmetric, the vertex must have an x-value located halfway between 0 and 1. Thus, the x-value of the vertex is 1/2. To find the y-value of the vertex, we evaluate f(1/2).

f(1/2) = –(1/2)² + (1/2) – 1 = –1/4 + (1/2) – 1 = –3/4.

Thus, the vertex of f(x) would be located at (1/2, –3/4), which is below the x-axis. Also, because f(0) and f(1) are below the vertex, we know that the parabola opens downward, and the focus must be below the vertex.

To summarize, we have just provided an example in which b is greater than zero, where f(x) has a vertex below the x-axis and a focus below the vertex. In other words, it is possible for b > 0, so it is not true that b must be less than 0.  

Let's go back to the original question. We know that a and c are both less than zero, so we know choices I and III must be true; however, we have just shown that b doesn't necessarily have to be less than zero. In other words, only I and III (but not II) must be  true. 

The answer is I and III only.

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.