Coordinate Geometry - 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 .

We are told that g(x) is found by taking f(x) and shifting it to the left by three and then down by four. This means that we can represent g(x) as follows:

g(x) = f(x + 3) - 4

Remember that the function f(x + 3) represents f(x) after it has been shifted to the LEFT by three, whereas f(x - 3) represents f(x) after being shifted to the RIGHT by three.

f(x) = -2x2 + 3x - 5

g(x) = f(x + 3) - 4 = \[-2(x+3)2 + 3(x+3) - 5\] - 4

g(x) = -2(x2 + 6x + 9) + 3x + 9 - 5 - 4

g(x) = -2x2 -12x -18 + 3x + 9 - 5 - 4

g(x) = -2x2 - 9x - 18 + 9 - 5 - 4

g(x) = -2x2 - 9x - 18

The answer is -2x2 - 9x - 18.

In order to reflect a function across the y-axis, all of the x-coordinates of every point on that function must be multiplied by negative one. However, the y-values of each point on the function will not change. Thus, we can represent the reflection of f(x) across the y-axis as f(-x). The figure below shows a generic function (not f(x) given in the problem) that has been reflected across the y-axis, in order to offer a better visual understanding.

Therefore, g(x) = f(–x).

f(x) = x3 – 2x2 + x – 4

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

g(x) = (–1)3x3 –2(–1)2x2 – x + 4

g(x) = –x3 –2x2 –x + 4.

The answer is –x3 –2x2 –x + 4.

When a point is reflected across the line y = x, the x and y coordinates are switched. In other words, the point (a, b) reflected across the line y = x would be (b, a).

Thus, if line m is reflected across the line y = x, the points that it passes through will be reflected across the line y = x. As a result, since m passes through (–4, 3) and (2, –6), when m is reflected across y = x, the points it will pass through become (3, –4) and (–6, 2).

Because line q is a reflection of line m across y = x, q must pass through the points (3, –4) and (–6, 2). We know two points on q, so if we determine the slope of q, we can then use the point-slope formula to find the equation of q.

First, let's find the slope between (3, –4) and (–6, 2) using the formula for slope between the points (x1, y1) and (x2, y2).

slope = (2 – (–4))/(–6 –3)

= 6/–9 = –2/3

Next, we can use the point-slope formula to find the equation for q.

y – y1 = slope(x – x1)

y – 2 = (–2/3)(x – (–6))

Multiply both sides by 3.

3(y – 2) = –2(x + 6)

3y – 6 = –2x – 12

Add 2x to both sides.

2x + 3y – 6 = –12

Add six to both sides.

2x + 3y = –6

The answer is 2x + 3y = –6.

f(x) undergoes a series of three transformations. The first transformation is the reflection of f(x) across the x-axis. This kind of transformation takes all of the negative values and makes them positive, and all of the positive values and makes them negative. This can be represented by multiplying f(x) by –1. Thus, –f(x) represents f(x) after it is reflected across the x-axis.

Next, the function is shifted to the left by four. In general, if g(x) is a function, then g(x – h) represents a shift by h units. If h is positive, then the shift is to the right, and if h is negative, then the shift is to the left. In order, to shift the function to the left by four, we would need to let h = –4. Thus, after –f(x) is shifted to the left by four, we can write this as –f(x – (–4)) = –f(x + 4).

The final transformation requires shifting the function down by 5. In general, if g(x) is a function, then g(x) + h represents a shift upward if h is positive and a shift downward if h is negative. Thus, a downward shift of 5 to the function –f(x + 4) would be represented as –f(x + 4) – 5.

The three transformations of f(x) can be represented as –f(x + 4) – 5.

The answer is –f(x + 4) – 5.

By definition, two angles form a linear pair if and only if

(1) they have the same vertex;

(2) they share a side; and,

(3) their interiors have no points in common.

In the figure below, and are marked in green and red, respectively:

The two angles have the same vertex and share no interior points. However, they do not share a side. Therefore, they do not comprise a linear pair.

A reflection over the line involves a switching of the coordinates to get us (7, 6). A reflection over the x-axis involves a negation of the y-coordinate. Thus the resulting point is (7, –6).

The coordinate length between you and your school is equivalent to the hypotenuse of a right triangle with sides of 3 and 4 units:

The distance is 5 coordinate lengths, and each coordinate length corresponds to 1.3 miles of distance, so

The amplitude is always the number in front of the trig function .

The general equation to remember is,

where

is amplitude

is period of and

is the phase shift.

In this case it's .

Two rays are opposite rays, by definition, if

(1) they have the same endpoint, and

(2) their union is a line.

The first letter in the name of a ray refers to its endpoint; the second refers to the name of any other point on the ray. and both have endpoint , so the first criterion is met. passes through point and passes through point ; and are indicated below in green and red, respectively:

The union of the two rays is a line. Both criteria are met, so the rays are indeed opposite.

The general formula for a circle with center (h,k) and radius r is .

The center of the original circle, therefore, is (2, -4).

If we move the circle to the left 2 spaces and down 3 spaces, then the center of the new circle is given by or .

If Jenny moves Bobby's circle up 2 units and to the right 1 unit, then the center of her circle is (3, 7). The radius remains 10.

The general equation for a circle with center (h, k) and radius r is given by

For Jenny's circle, (h, k) = (3, 7) and r=10.

Substituting these values into the general equation gives us