The graph has x-intercepts at x = 0 and x = 8. This indicates that 0 and 8 are roots of the function.

The function must take the form y = x(x - 8) in order for these roots to be true.

The parabola opens downward, indicating a negative leading coefficient. Expand the equation to get our answer.

y = -x(x - 8)

y = -x2 + 8x

y = 8x - x2

Therefore, the answer must be y = 8x - x2

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 .

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 .

This question relies on both the vertical-line test and the definition of a function. We need to use the vertical-line test to determine which of the graphs is not a function (i.e. the graph that has more than one output for a given input). The vertical-line test states that a graph represents a function when a vertical line can be drawn at every point in the graph and only intersect it at one point; thus, if a vertical line is drawn in a graph and it intersects that graph at more than one point, then the graph is not a function. The circle is the only answer choice that fails the vertical-line test, and so it is not a function.

A line is an asymptote in a graph if the graph of the function nears the line as X or Y gets larger in absolute value.

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.

There are four shifts of the graph y = f(x):

y = f(x) + c shifts the graph c units upwards.

y = f(x) – c shifts the graph c units downwards.

y = f(x + c) shifts the graph c units to the left.

y = f(x – c) shifts the graph c units to the right.

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.

The denominator cannot be zero, otherwise the function is indefinite. Therefore x cannot be –2 or –3.

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: