How to find transformation for an analytic geometry equation - Geometry

Regular pentagons have lines of symmetry through each vertex and the center of the opposite side, meaning the y-axis forms a line of symmetry in this instance. Therefore, point P is negative b units in the x-direction, and c units in the y-direction. It is a reflection of point (b,c) across the y-axis.

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.

The period is the time it takes for the graph to complete one cycle.

In this particular case we have a sine curve that starts at 0 and completes one cycle when it reaches .

Therefore, the period is

The phase displacement is the shift from the center of the graph. Since this is a sine graph and the sin(0) = 0, this is in phase.

The phase displacement is the shift of the graph. Since cos(0) = 1, the phase shift is π because the graph is at its high point then.

To solve this question, you must have an understanding of standard transformations. To move a function along the x-axis in the positive direction, you must subtract the value from the operative x-value. For example, to move a function, f(x), five units to the left would be f(x+5).

To shift a function along the y-axis in the positive direction, you must add the value to the overall function. For example, to move a function, f(x), three units up would be f(x)+3.

The question asks us to move the function, f(x), left three units and up four units. f(x+3) will move the function three units to the left and f(x)+4 will move it four units up.

Together, this gives our final answer of f(x+3)+4.

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 y=x 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

When a function is flipped across the x-axis, the new function is equal to . Therefore, our function is equal to:

Our final answer is therefore

When a function is flipped across the y-axis, the resulting function is equal to . Therefore, to find our , we must substitute in for every is our equation:

Our final answer is therefore

When a function is transformed units upwards, the new function is equal to . Likewise, if is transformed units to the right, the new function is equal to . Therfore, we can first find the upwards transformation by adding to the function:

Now we can apply the horizontal transformation by replacing all 's in the function with . Our transformed function therefore becomes:

We then multiply this out to obtain:

Our final answer is therefore

When we reflect a point across the line, , we swap the x- and y-coordinates; therefore, in each point, we will switch the x and y-coordinates:

becomes ,

becomes , and

becomes .

The correct answer is the following:

The other answer choices are incorrect because we only use the negatives of the coordinate points if we are flipping across either the x- or y-axis.

The standard form of a linear equation is Here, we are given two equations, and , which differ only in their terms. In other words, these functions differ only in their slope. has a larger slope than does , so is steeper.

Algebraic transformations of functions rely on manipulating components of the equation's standard form. The standard form of a linear equation is . Changes to the slope () will make the graph steeper or shallower, changes to the y-intercept () will shift the graph vertically, and changes to the indepent variable () will shift the graph horizontally. Here, we are given and asked to transform it into a new equation vertically shifted up four units. We can accomplish this by adding four to the constant term, so the correct answer is .

multiples the slope by four, which will result in a steeper graph.

subtracts four from the constant term, which shifts the graph vertically, but in the wrong direction.

adds four to the independent variable, which shifts the graph horizontally to the left.

Algebraic transformations of functions rely on manipulating components of the equation's standard form. The standard form of a linear equation is . Changes to the slope () will make the graph steeper or shallower, changes to the y-intercept () will shift the graph vertically, and changes to the indepent variable () will shift the graph horizontally. Here, we are given and asked to transform it into a new equation horizontally shifted to the right two units. We can accomplish this by subtracting two from the independent variable, so the correct answer is .

adds two to the constant term, which shifts the graph vertically.

adds two to the independent variable which shifts the graph horizontally, but in the wrong direction.

multiplies the slope by two, which makes the graph steeper.

Algebraic transformations of functions rely on manipulating components of the equation's standard form. The standard form of a linear equation is . Changes to the slope () will make the graph steeper or shallower, changes to the y-intercept () will shift the graph vertically, and changes to the indepent variable () will shift the graph horizontally. Here, we are given and asked to transform it into a new equation that increases the slope by three, shifts the graph horizontally one unit to the left, and shifts the graph vertically three units down.

First, multiply the slope by three.

Add one to the independent variable.

Subtract three from the constant term.

correctly increases the slope and subtracts three from the constant term, but fails to properly substitute for , leading to an erroneous simplification. In other words, becomes .

shifts the function to the right instead of the left.

shifts the function both right and up, rather than left and down.