Graphing - ACT Math

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.

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.

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

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.

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

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

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:

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 .

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 .

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 .

First, we need to find the slope of the above line.

The slope of a line. given two points can be calculated using the slope formula

Set :

The slope of a line perpendicular to it has as its slope the opposite of the reciprocal of 2, which would be . Since we want this line to have the same -intercept as the first line, which is the point , we can substitute and in the slope-intercept form:

Plug in for to find a point on the line:

Thus, is a point on the line.

Plug in for to find a second point on the line:

is another point on the line.

Now we know that the line passes through the points and .

A quick sketch of the two points reveals that the line passes through all but the third quadrant.

The distance formula is given by d = square root \[(x2 – x1)2 + (y2 – y1)2\]. Let point 2 be (7,13) and point 1 be (1,5). Substitute the values and solve.

The x-coordinate for the midpoint is given by taking the arithmetic average (mean) of the x-coordinates of the two end points. So the x-coordinate of the midpoint is given by

The same procedure is used for the y-coordinates. So the y-coordinate of the midpoint is given by

Thus the midpoint is given by the ordered pair

The slope is found using the formula .

We know that the line contains the points (3,0) and (0,6). Using these points in the above equation allows us to calculate the slope.

is the -intercept and equals . can be solved for by substituting in the equation for , which yields

The amplitude is half the measure from a trough to a peak.

One way to answer this is to first find the equation of the line.

The slope of a line. given two points can be calculated using the slope formula

Set :

The line has slope 3 and -intercept , so we can substitute in the slope-intercept form:

Now substitute 4 for and for and solve for :