An equation of a line is made of two parts: a slope and a y-intercept.

The y-intercept is where the function crosses the y-axis which in this problem it is 0.

The slope is determined by the rise of the function over the run which is  , so the function is moving up one and over one.

Therefore your equation is:

, which is simply

The first number,  indicates how far the point is positioned to the left or right of the origin. Because the number is negative, the point is three units to the LEFT of the origin. The second number indicates how far the point is postioned up or down from the origin. Because the number is positive, the point is located four units above the origin.

To answer this question, we will have to manipulate the distance formula:

To get rid of the square root, we can square both sides:

and plug in the information given in the question.

At this point we can simply plug in the possible values to determine which combination of coordinates will make the equation above true. 

Thus the correct coordinate is,

.

To graph these points we just need to remember that the first number is the x value and the second number is the y value. For (A) we have (1,3). So we move one tick over on the positive x-axis. Then from there we move up to the third tick on the y axis:

If the value is negative we must move in the other direction. So, for all 4 points,

When trying to determine coordinates of a point you need to look at the value of x first (how many units left or right the point is, then the y-value (how many up or down it is).

When you look at this point you see that it is moved right 2 units and up 1.

So your coordinates are:

A function may only have one y-value for each x-value.

The vertical line test can be used to identify the function. If at any point on the graph, a straight vertical line intersects the curve at more than one point, the curve is not a function.

As is clear from the graph, in the interval between  ( included) to , the  is constant at  and then from ( not included) to  ( not included), the  is a decreasing function.

We have the following answer choices.

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.

The asymptote of this equation can be found by observing that  regardless of . We are thus solving for the value of as approaches zero.

So the value that  cannot exceed is , and the line  is the asymptote.

An exponential equation of the form  has only one asymptote - a horizontal one at . In the given function, , so its one and only asymptote is .