Because points on a graph are written in the form of, and the point given was, this means thatand.

In order to solve for, these values forandmust be plugged into the given equation. This gives us the following:

We then solve the equation by finding the value of the right side, then dividing the entire equation by 5, as follows:

Therefore, the value ofis.

一个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.

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

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

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

然而,图表显示,f(1) = 4。作为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 graph is a down-opening parabola with a maximum of. Therefore, there are no y values greater than this for this function.

一个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.