One of the properties of taking an absolute value of a function is that the values are all made positive. The values themselves do not change; only their signs do. In this graph, none of the y-values are negative, so none of them would change. Thus the two graphs should be identical.

First, because the graph consists of pieces that are straight lines, the function must include an absolute value, whose functions usually have a distinctive "V" shape. Thus, we can eliminate f(x) = x² – 4x + 3 from our choices. Furthermore, functions with x² terms are curved parabolas, and do not have straight line segments. This means that f(x) = |x² – 4x| – 3 is not the correct choice. 

Next, let's examine f(x) = |2x – 6|. Because this function consists of an abolute value by itself, its graph will not have any negative values. An absolute value by itself will only yield non-negative numbers. Therefore, because the graph dips below the x-axis (which means f(x) has negative values), f(x) = |2x – 6| cannot be the correct answer. 

Next, we can analyze f(x) = |x – 1| – 2. Let's allow x to equal 1 and see what value we would obtain from f(1). 

f(1) = | 1 – 1 | – 2 = 0 – 2 = –2

However, the graph above shows that f(1) = –4. As a result, f(x) = |x – 1| – 2 cannot be the correct equation for the function. 

By process of elimination, the answer must be f(x) = |2x – 2| – 4. We can verify this by plugging in several values of x into this equation. For example f(1) = |2 – 2| – 4 = –4, which corresponds to the point (1, –4) on the graph above. Likewise, if we plug 3 or –1 into the equation f(x) = |2x – 2| – 4, we obtain zero, meaning that the graph should cross the x-axis at 3 and –1. According to the graph above, this is exactly what happens. 

The answer is f(x) = |2x – 2| – 4.

A line has the equation

 where  is the  intercept and  is the slope.

The  intercept can be found by noting the point where the line and the y-axis cross, in this case, at  so .

The slope can be found by selecting two points, for example, the y-intercept and the next point over that crosses an even point, for example, .

Now applying the slope formula,

 which yields .

Therefore the equation of the line becomes:

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.