图为down-opening抛物线极限m of. 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 x²terms 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

whereis theintercept andis the slope.

Theintercept can be found by noting the point where the line and the y-axis cross, in this case, atso.

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 ofwhen.

Here, we are given the function. In order to calculate the y-intercept, setequal 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 offor which.

Here, we are given the function. In order to calculate the x-intercept, setequal 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 fromto, is the furthest from home. So the answer will be fromto.