The graph is a down-opening parabola with a maximum of \(\displaystyle y=3\). Therefore, there are no y values greater than this for this function.

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

\(\displaystyle y=mx+b\) where \(\displaystyle b\) is the \(\displaystyle y\) intercept and \(\displaystyle m\) is the slope.

The \(\displaystyle y\) intercept can be found by noting the point where the line and the y-axis cross, in this case, at \(\displaystyle (0,2)\) so \(\displaystyle b=2\).

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, \(\displaystyle (2, -1)\).

Now applying the slope formula,

 \(\displaystyle m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

 which yields \(\displaystyle m=\frac{-1-2}{2-0}=-\frac{3}{2}\).

Therefore the equation of the line becomes:

\(\displaystyle y=-\frac{3}{2}x+2\)

Graphically, the y-intercept is the point at which the graph touches the y-axis.  Algebraically, it is the value of \(\displaystyle y\) when \(\displaystyle x=0\).

Here, we are given the function \(\displaystyle f(x)=2x+4\).  In order to calculate the y-intercept, set \(\displaystyle x\) equal to zero and solve for \(\displaystyle y\).

\(\displaystyle y=2(0)+4\)

\(\displaystyle y=4\)

So the y-intercept is at \(\displaystyle (0,4)\).

Graphically, the x-intercept is the point at which the graph touches the x-axis.  Algebraically, it is the value of \(\displaystyle x\) for which \(\displaystyle y=0\).

Here, we are given the function \(\displaystyle f(x)=2x+4\).  In order to calculate the x-intercept, set \(\displaystyle y\) equal to zero and solve for \(\displaystyle x\).

\(\displaystyle 0=2x+4\)

\(\displaystyle -4=2x\)

\(\displaystyle -2=x\)

So the x-intercept is at \(\displaystyle (-2,0)\).

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.

\(\displaystyle f(x)=\frac{1}{2}x-2\)

Let \(\displaystyle x=0\).  Then

\(\displaystyle y=\frac{1}{2}(0)-2\)

\(\displaystyle y=-2\)

So our first set of points (which is also the y-intercept) is \(\displaystyle (0,-2)\)

Let \(\displaystyle y=0\).  Then

\(\displaystyle 0=\frac{1}{2}x-2\)

\(\displaystyle 2=\frac{1}{2}x\)

\(\displaystyle 4=x\)

So our second set of points (which is also the x-intercept) is \(\displaystyle (4,0))\).

If we look at the graph, the line segment from \(\displaystyle (8,15)\) to \(\displaystyle (14,15)\), is the furthest from home. So the answer will be from \(\displaystyle \text{Time}=8\) to \(\displaystyle \text{Time}=14\).

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 \(\displaystyle (12, 0 )\) and \(\displaystyle (0,12)\) can be found using the slope formula, setting \(\displaystyle x_{1} = 12 , y_{2} = x_{2} = 0, y_{2} = 12\):

\(\displaystyle m_{1} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{12-0}{0-12} = \frac{12}{-12} = -1\)

The slope of the line that connects \(\displaystyle (12, 0 )\) and \(\displaystyle (0,-12)\) can be found similarly, setting \(\displaystyle x_{1} = 12 , y_{2} = x_{2} = 0, y_{2} = -12\):

\(\displaystyle m_{2} = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}= \frac{-12-0}{0-12} = \frac{-12}{-12} = 1\)

The product of their slopes is \(\displaystyle m_{1} m_{2} = -1 \cdot 1 = -1\), which indicates perpendicularity between the sides.

This makes the triangle right, and the side with endpoints \(\displaystyle (12, 0 )\) and \(\displaystyle (0,-12)\) 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, \(\displaystyle (0,0)\).