All Algebra II Resources
Example Questions
Example Question #11 : Transformations
If the function is linear and the function is quadratic, then the function is?
not a function
quadratic
a linear quadratic function
linear function
higher order polynomial function
quadratic
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.
Example Question #11 : Transformations
Reflect across the x-axis, then reflect across , and then shift this line up five units. What is the new equation?
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:
Example Question #191 : Functions And Graphs
Reflect the line across the line , and then reflect again across the line . What is the new equation of this horizontal line?
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:
Example Question #15 : Transformations
If the function is shifted down two units and left four units, what is the new y-intercept?
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 .
Example Question #13 : Transformations
Reflect the line across , and shift the line down three units. What is the new equation?
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:
Example Question #14 : Transformations
Give the equation of the vertical asymptote of the graph of the equation .
The graph of has no vertical asymptote.
The graph of has no vertical asymptote.
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.
Example Question #15 : Transformations
Give the equation of the horizontal asymptote of the graph of the equation
The graph of has no horizontal 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.
Example Question #11 : Transformations
Shift down three units. What is the new equation?
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:
Example Question #1 : Inverse Functions
Which of the following represents ?
The question is asking for the inverse function. To find the inverse, first switch input and output -- which is usually easiest if you use notation instead of . Then, solve for .
Here's where we switch:
To solve for , we first have to get it out of the denominator. We do that by multiplying both sides by .
Distribute:
Get all the terms on the same side of the equation:
Factor out a :
Divide by :
This is our inverse function!
Example Question #2 : Inverse Functions
What is the inverse of the following function?
Let's say that the function takes the input and yields the output . In math terms:
So, the inverse function needs to take the input and yield the output :
So, to answer this question, we need to flip the inputs and outputs for . We do this by replacing with (or a dummy variable; I used ) and with . Then we solve for to get our inverse function:
Now we solve for by subtracting from both sides, taking the cube root, and then adding :
is our inverse function,
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