Graphing - SAT Math

The graph is a down-opening parabola with a maximum 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 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.

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 .

If we look at the graph, the line segment from to , is the furthest from home. So the answer will be from to .

Using the distance formula, calculate the distance from each of these points to the origin, (0, 0). While each answer choice has coordinates that add up to seven, (-1, 8) is the coordinate pair that produces the largest distance, namely , or approximately 8.06.

On the coordinate plane, a point with a negative -coordinate and a positive -coordinate lies in the upper left quadrant—Quadrant II.

Another way of viewing this problem is to note that the three given vertices form a triangle whose sides' perpendicular bisectors intersect at the points , , and . However, the three perpendicular bisectors of the sides of any triangle always intersect at a common point. The correct response is that , , and are the same point.

The referenced figure is below.

The two non-horizontal line segments are perpendicular, as is proved as follows:

The slope of the line that connects and can be found using the slope formula, setting :

The slope of the line that connects and can be found similarly, setting :

The product of their slopes is , which indicates perpendicularity between the sides.

This makes the triangle right, and the side with endpoints and the hypotenuse. The center of the circle that circumscribes a right triangle is the midpoint of its hypotenuse, which is easily be seen to be the origin, .

The referenced figure is below.

The triangle formed is a right triangle whose hypotenuse is the segment with the endpoints and . The center of the circle that circumscribes a right triangle is the midpoint of its hypotenuse, so the midpoint formula

can be applied, setting :

The midpoint of the hypotenuse, and, consequently, the center of the circumscribed circle, is the point with coordinates .

The length of the horizontal leg of the triangle is the distance from the origin to , which is 8.

The area of a right triangle is half the product of the lengths of its legs and , so, setting and and solving for :

Therefore, the length of the vertical leg is 10, and, since the -intercept of the line containing the hypotenuse is on the positive -axis, this intercept is . The slope of a line with intercepts is

,

so, setting and :

Set and in the slope-intercept form of the equation of a line,

;

the line has equation

The -coordinate of the point on the line with -coordinate 2 can be found using substitution; setting ::

Mrs. Smith's 8th grade class had a quiz last week. The graph below depicts the number of questions students got incorrect on their quiz and their corresponding quiz grade. In other words, the graph in this particular question is a dot plot and the question asks to find a correlation if one exists.

Recall that a correlation is a trend seen in the data. Graphically, trends can be either:

I. Positive

II. Negative

III. Constant

IV. No trend

For a trend to be positive the x and y variable both increase. A trend is negative when the y variable (dependent variable) decreases as the x variable (independent variable) increases. A constant trend occurs when the y variable stays the same as the x variable increases. No trend exists when the data appears to be scattered with no association between the x and y variables.

Examining the graph given it is seen that the x variable is the number of questions missed and the y variable is the overall score on the quiz. It is seen that as the number of questions missed increases, the overall score on the quiz decreases. This describes a negative trend.

In other words, the graph depicts a negative correlation.

Given the graph of record sales, to find the fraction of records that were sold in 2004 to 2010 first identify the record sales in 2004 and the record sales in 2010.

Examining the graph,

Record sales in 2004:14 million

Record sales in 2010: 13 million

From here, to find the fraction of records sold during this time period, use the following formula.

Let l = length and w = width of the pen. Let us assume that the side blocked by the mountain is along the length of the pen.

The length of wire used to make the pen must equal l + 2w, because this is the perimeter of a rectangle, excluding one of the lengths. The area of the pen will equal l x w.

l + 2w = 20

l = 20-2w

A = l x w = (20-2w)(w) = 20w - 4w2

Let A be a function of w, such that A(w) = 20w - 4w2. We want to find the maximum value of A. We recognize that the graph of A must be in the shape of a parabola, pointing downward. The maximum value of the parabola will thus occur at the vertex.

We want to rewrite A(w) in the standard form of a parabola, given by f(x) = a(x-h)2+k. In order to do this, we must complete the square.

20w-4w2 = -4w2+20w = -4(w2-5w) = -4(w2-5w + 25/4) + 25 = -4(w-5/2)2+25

Thus, the vertex of the parabola occurs at (5/2, 25), which means that w = 5/2.

Going back to our original equation, l + 2(5/2) = 20, and l = 15.

A = l x w = 15(5/2) = 75/2

The function is given in vertex form, which is (x-h)2+k where the vertex of the parabola is the point (h,k). In this particular function, h=-4 and k=3, so the vertex is (-4,-3). No parabola is one-to-one, as they don't pass the horizontal line test. While parabolas can be even functions, this will only happen when the vertex is on the y-axis (or when h=0) because even functions must be symmetric with respect to the y-axis. No function can have two y-intercepts, as it would then not pass the vertical line test and not be a function. This parabola does have two x-intercepts, however. This can be shown by setting y=0 and solving for x, or by simply realizing that the vertex is below the x-axis and the parabola opens up.

Because the graph of f(x) = x2 is symmetric about the y-axis, when we shift it downward, the points where it intersects the x-axis will be the same distance from the origin. In other words, we could say that one intercept will be (-a,0) and the other would be (a,0). The distance between these two points has to be 8, so that means that 2a = 8, and a = 4. This means that when f(x) is shifted downward, its new roots will be at (-4,0) and (4,0).

Let g(x) be the graph after f(x) has been shifted downward. We know that g(x) must have the roots (-4,0) and (4,0). We could thus write the equation of g(x) as (x-(-4))(x-4) = (x+4)(x-4) = x2 - 16.

We can now compare f(x) and g(x), and we see that g(x) could be obtained if f(x) were shifted down by 16 units; therefore, the answer is 16.

f(x) must be a parabola, since it contains an x2 term. We are told that the vertex is below the x-axis, and that the focus is below the vertex. Because a parabola always opens toward the focus, f(x) must point downward. The general graph of the parabola must have a shape similar to this:

Since the parabola points downward, the value of a must be less than zero. Also, since the parabola points downward, it must intersect the y-axis at a point below the origin; therefore, we know that the value of the y-coordinate of the y-intercept is less than zero. To find the y-coordinate of the y-intercept of f(x), we must find the value of f(x) where x = 0. (Any graph intersects the y-axis when x = 0.) When x = 0, f(0) = a(0) + b(0) + c = c. In other words, c represents the value of the y-intercept of f(x), which we have already established must be less than zero. To summarize, a and c must both be less than zero.

The last number we must analyze is b. One way to determine whether b must be negative is to assume that b is NOT negative, and see if f(x) still has a vertex below the x-axis and a focus below the vertex. In other words, let's pretend that b = 1 (we are told b is not zero), and see what happens. Because we know that a and c are negative, let's assume that a and c are both –1.

If b = 1, and if a and c = –1, then f(x) = –x2 + x – 1.

Let's graph f(x) by trying different values of x.

If x = 0, f(x) = –1.

If x = 1, f(x) = –1.

Because parabolas are symmetric, the vertex must have an x-value located halfway between 0 and 1. Thus, the x-value of the vertex is 1/2. To find the y-value of the vertex, we evaluate f(1/2).

f(1/2) = –(1/2)2 + (1/2) – 1 = –1/4 + (1/2) – 1 = –3/4.

Thus, the vertex of f(x) would be located at (1/2, –3/4), which is below the x-axis. Also, because f(0) and f(1) are below the vertex, we know that the parabola opens downward, and the focus must be below the vertex.

To summarize, we have just provided an example in which b is greater than zero, where f(x) has a vertex below the x-axis and a focus below the vertex. In other words, it is possible for b > 0, so it is not true that b must be less than 0.

Let's go back to the original question. We know that a and c are both less than zero, so we know choices I and III must be true; however, we have just shown that b doesn't necessarily have to be less than zero. In other words, only I and III (but not II) must be true.

The answer is I and III only.