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 .

If we look at the graph, the line segment from  to , is the furthest from home. So the answer will be from  to .

The referenced figure is below. 

The two non-horizontal line segments are perpendicular, as is proved as follows:

The slope of the line that connects  and  can be found using the slope formula, setting :

The slope of the line that connects  and  can be found similarly, setting :

The product of their slopes is , which indicates perpendicularity between the sides.

This makes the triangle right, and the side with endpoints  and  the hypotenuse. The center of the circle that circumscribes a right triangle is the midpoint of its hypotenuse, which is easily be seen to be the origin, .

The referenced figure is below. 

The triangle formed is a right triangle whose hypotenuse is the segment with the endpoints  and . The center of the circle that circumscribes a right triangle is the midpoint of its hypotenuse, so the midpoint formula 

can be applied, setting :

The midpoint of the hypotenuse, and, consequently, the center of the circumscribed circle, is the point with coordinates .

Another way of viewing this problem is to note that the three given vertices form a triangle  whose sides' perpendicular bisectors intersect at the points , , and . However, the three perpendicular bisectors of the sides of any triangle always intersect at a common point. The correct response is that , , and  are the same point.

The length of the horizontal leg of the triangle is the distance from the origin  to , which is 8.

The area of a right triangle is half the product of the lengths of its legs  and , so, setting  and  and solving for :

Therefore, the length of the vertical leg is 10, and, since the -intercept of the line containing the hypotenuse is on the positive -axis, this intercept is . The slope of a line with intercepts  is 

, 

so, setting  and :

Set  and  in the slope-intercept form of the equation of a line, 

;

the line has equation

The -coordinate  of the point on the line with -coordinate 2 can be found using substitution; setting ::