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

Our final answer is therefore

When a function是翻across the y-axis, the resulting functionis equal to. Therefore, to find our, we must substitute infor everyis our equation:

Our final answer is therefore

When a functionis transformedunits upwards, the new functionis equal to. Likewise, ifis transformedunits to the right, the new functionis equal to. Therfore, we can first find the upwards transformation by addingto 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

The standard form of a linear equation isHere, we are given two equations,and, which differ only in theirterms. In other words, these functions differ only in their slope.has a larger slope than does, sois 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 givenand 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 givenand 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 givenand 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 substitutefor, 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 isπbecause 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 negativebunits in the x-direction, andcunits in the y-direction. It is a reflection of point (b,c) across the y-axis.