Where will the point  be located after the following transformations?

1. Reflection about the x-axis results in multiplying the y value by negative one thus .
2. Translation up 3, means to add three to the y values which results in  .
3. Translation right 4, means to add four to the x value which will result in .

The transformation for a left shift along the x-axis for requires we add  to the argument of the function .  

The y-intercept of the linear function  is .  

The linear function  will have the form, 

Where  is the y-intercept and  is the slope; both are constant. 

The quadratic function  will have the form, 

We are given that the function  is defined, 

we obtain another function that is also a quadratic function since  and  are constants. Therefore,  is quadratic. 

Reflect  across the x-axis will turn the equation to:

If we then reflect  across , the equation will become:

Shifting this line up five units means that we will add five to this equation.

The equation after all the transformations is:  

The answer is:  

The distance between  and  is three units.  If the line  is reflected across , this means that the new line will also be three units away from .

The equation of the line after this reflection is:  

If this line is reflected again across the line , both lines are six units apart, and the reflection would mean that the line below line  is also six units apart.

Subtract six from line .

The equation of the line after the transformations is:  

The answer is:  

If the graph was shifted two units down, only the y-intercept will change, and will decrease by two.

The new equation is:  

If the graph was shifted left four units, the root will shift four units to the left, and the  will need to be replaced with .

The new y-intercept will be .

The equations with an existing  variable is incorrect because they either represent lines with slopes or vertical lines.

After the line  is reflected across , the line becomes .

Shifting this line down three units mean that the line will have a vertical translation down three.

Subtract the equation  by three.

The result is:  

Define . As an exponential function, this has a graph that has no vertical asymptote, as  is defined for all real values of .  In terms of  :

,

The graph of  is a transformation of that of  - a horizontal shift (  ), a vertical stretch (  ), and a vertical shift (  ) of the graph of ; none of these transformations changes the status of the function as one whose graph has no vertical asymptote.

Define  in terms of , 

It can be restated as the following:

The graph of  has as its horizontal asymptote the line of the equation . The graph of  is a transformation of that of  —a right shift of 2 units  , a vertical stretch   , and an upward shift of 5 units  . The right shift and the vertical stretch do not affect the position of the horizontal asymptote, but the upward shift moves the asymptote to the line of the equation . This is the correct response.

The equation is currently in standard form.  Rewrite the current equation in slope-intercept form.  Subtract  from both sides.

Since the graph is shifted downward three units, all we need to change is decrease the y-intercept by three.  Subtract the right side by three.

The answer is: