How to find transformation for an analytic geometry equation - SAT Math

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.

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 .

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

A plane can be named after any three points on the plane that are not on the same line. As seen below, points , , and are on the same line.

Therefore, Plane is not a valid name for the plane.

A line can be named after any two points it passes through. The line is indicated in green below.

The line does not pass through , so cannot be part of the name of the line. Specifically, is not a valid name.

By definition, two angles comprise a pair of vertical angles if

(1) they have the same vertex; and

(2) the union of the two angles is exactly a pair of intersecting lines.

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

While the two angles have the same vertex, their union is not a pair of intersecting lines. The two angles are not a vertical pair.

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.

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.