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.

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.

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.

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.

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.

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.