Graphing - PSAT Math

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 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 .

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 slope of the line is and the y-intercept is .

The equation of the line is .

This is the parent graph of . Since the graph in question is negative, then we flip the quadrants in which it will approach infinity. So the graph of will start in quadrant 2 and end in 4.

In order to find the equation of a line in slope-intercept form , where is the slope and is the y-intercept), one must know or otherwise figure out the slope of the line (its rate of change) and the point at which it intersects the y-axis. By looking at the graph, you can see that the line crosses the y-axis at . Therefore, .

Slope is the rate of change of a line, which can be calculated by figuring out the change in y divided by the change in x, using the formula

.

When looking at a graph, you can pick two points on a graph and substitute their x- and y-values into that equation. On this graph, it's easier to choose points like and . Plug them into the equation, and you get

Plugging in those values for and in the equation, and you get

Answer: (1/2,0) and (0,-2)

Finding the y-intercept: The y-intercept is the point at which the line crosses tye y-axis, meaning that x = 0 and the format of the ordered pair is (0,y) with y being the y-intercept. The equation is in slope-intercept () form, meaning that the y-intercept, b, is actually given in the equation. b = -2, which means that our y-intercept is -2. The ordered pair for expressing this is (0,-2)

Finding the x-intercept: To find the x-intercept of the equation , we must find the point where the line of the equation crosses the x-axis. In other words, we must find the point on the line where y is equal to 0, as it is when crossing the x-axis. Therefore, substitute 0 into the equation and solve for x:

The x-interecept is therefore (1/2,0).

The line in the diagram has a negative slope and a positive y-intercept. It has a negative slope because the line moves from the upper left to the lower right, and it has a positive y-intercept because the line intercepts the y-axis above zero.

The only answer choice with a negative slope and a positive y-intercept is

The point shifted to the right 5 units will shift along the x-axis, meaning that you will add 5 to the original x-coordinate, so the new . The point shifted down by three units will shift down the y-axis, meaning that you will subtract three from the original y-coordinate, so the new .

The resultant coordinate is .

The point can be reached from the origin by moving 2 units right then 6 units up. This makes the first coordinate 2 and the second coordinate 6.

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.

Because we are given the vertex of the parabola, the easiest way to solve this problem will involve the use of the formula of a parabola in vertex form. The vertex form of a parabola is given by the following equation:

f(x) = a(x – h)2 + k, where (h, k) is the location of the vertex, and a is a constant.

Since the parabola has its vertex as (–3, 4), its equation in vertex form must be as follows:

f(x) = a(x – (–3)2 + 4 = a(x + 3)2 + 4

In order to complete the equation for the parabola, we must find the value of a. We can use the point (–1, –4), through which the parabola passes, in order to determine the value of a. We can substitute –1 in for x and –4 in for f(x).

f(x) = a(x + 3)2 + 4

–4 = a(–1 + 3)2 + 4

–4 = a(2)2 + 4

–4 = 4_a_ + 4

Subtract 4 from both sides.

–8 = 4_a_

Divide both sides by 4.

a = –2

This means that the final vertex form of the parabola is equal to f(x) = –2(x + 3)2 + 4. However, since the answer choices are given in standard form, not vertex form, we must expand our equation for f(x) and write it in standard form.

f(x) = –2(x + 3)2 + 4

= –2(x + 3)(x + 3) + 4

We can use the FOIL method to evaluate (x + 3)(x + 3).

= –2(x_2 + 3_x + 3_x_ + 9) + 4

= –2(x_2 + 6_x + 9) + 4

= –2_x_2 – 12_x_ – 18 + 4

= –2_x_2 – 12_x_ – 14

The answer is f(x) = –2_x_2 – 12_x_ – 14.