All Algebra II Resources
Example Questions
Example Question #71 : Solving Absolute Value Equations
Solve.
and
and
or
or
or
or
Solve.
Step 1: Isolate the absolute value by subtracting from both sides of the equation.
Step 2: Divide -1 from both sides of the equation in order to get rid of the negative sign in front of the absolute value.
Step 3: Because this is an inequality, this equation can be solved in two parts as shown below.
Note: can be written as or
or
Step 4: Solve both parts.
Distribute the .
Subtract from both sides of the equation.
Divide both sides of the equation by .
Distribute the .
Subtract from both sides of the equation.
Divide both sides of the equation by .
Step 5: Combine both parts using "or".
or
Solution: or
Example Question #72 : Solving Absolute Value Equations
Solve for values given the equation
and
and
and
and
Given:
When given an absolute value recognize there are often multiple solutions. The reason why is best exhibited in a simpler example:
Given solve for values of that make this statement true. When you taken an absolute value of something you always end up with the positive number so both and would make this statement true. The solutions can also be written as .
In the case of the more complicated equation for the same reason there are potentially two solutions, which are shown by as an absolute value will always end up creating a positive result.
To simplify the absolute value we must look at each of these cases:
Let's start with the positive case:
Just like a normal equation with one unknown we will simplify it by isolating by itself. This is first done by subtracting from both sides leaving:
Next is divided from both sides leaving , as your final solution.
To check this solution it must be substituted in the original absolute value for and if it's a correct answer you'll end up with a true statement, so
,
so this becomes:
and the absolute value of is so you end up with a true statement. Therefore is a valid solution
Next let's solve for the negative case:
Distribute the negative sign, which is just to make calculations easier and you'll get:
Next can be added to both sides, giving
dividing by leaves:
Checking this solution is done just as you did for the previous solution obtained.
Given ,
substitute in for
multiply gives
so you obtain
adding
and the absolute value of is thereby making this also a valid solution, therefore the two valid solutions are and
Example Question #2197 : Mathematical Relationships And Basic Graphs
Solve for .
and
and
and
and
Solve for given
When given an absolute value equation recognize there are often multiple solutions. The reason why is best exhibited in a simpler example:
Given solve for values of that make this statement true. When you taken an absolute value of something you always end up with the positive number so both and would make this statement true. The solutions can also be written as ±.
In the case of the more complicated equation , however before proceeding let's simplify this equation a little as there are two terms that can be added together within the absolute value. These are and .
When added together this gives ,
thereby giving you
For the same reason as shown in the case of there are potentially two solutions to this equation, which are shown by as an absolute value will always end up creating a positive result. Make sure you apply the parentheses to only the portion of the equation with the absolute value otherwise your answer will be incorrect.
To simplify the absolute value we must look at each of these cases:
Let's start with the positive case:
Just like a normal equation with one unknown we will simplify it by isolating by itself. This is first done by combining like terms and getting on one side of the equation.
We can first subtract from which is , this gives you:
Next we can subtract from both sides of the equation.
Dividing both sides by gives you a final answer of this however can be simplified to as both the numerator and denominator are divisible by .
so
To check this solution it must be substituted in the original absolute value for and if it's a correct answer you'll end up with a true statement, so
,
Simplify the equation by multiplying by and by
This leaves our equation with
Next add to both sides of the equation:
In order to simplify you must find a common denominator, which is most easily done by multiplying .
This leaves:
Similarly a common denominator is found for and by multiplying by which gives you
this leaves:
simplifying further gives you
which is a true statement, so is a valid solution
Next let's solve for the negative case:
Distribute the negative sign, which is just to make calculations easier and you'll get:
Combine like terms:
This can be simplified to as both the numerator and denominator are divisible by , therefore you final answer is
Checking this solution is done just as you did for the previous solution obtained.
Given
substitute in for
Multiplying by gives
Multiplying gives
So the equation simplifies to:
Next can be added to both sides of the equation giving:
Now common denominators must be found for and
The common denominator for is found by multiplying by . This gives you . This can then be simplified through addition and gives .
The common denominator of is found by multiplying and
so
The simplified equation becomes
Through dividing by you get and the absolute value of is , so you get . This is true statement
so is also a valid solution.
Example Question #12 : Algebra Ii
Solve for :
To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.
and
This gives us:
and
However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as cannot equal a negative number.
Our final solution is then
Example Question #2201 : Mathematical Relationships And Basic Graphs
Which values of provide the full solution set for the inequality:
Example Question #1 : Graphing Absolute Value Functions
Refer to the above figure.
Which of the following functions is graphed?
Below is the graph of :
The given graph is the graph of reflected in the -axis, then translated up 6 units. This graph is
, where .
The function graphed is therefore
Example Question #1 : Graphing Absolute Value Functions
Refer to the above figure.
Which of the following functions is graphed?
Below is the graph of :
The given graph is the graph of reflected in the -axis, then translated left 2 units (or, equivalently, right units. This graph is
, where .
The function graphed is therefore
Example Question #2 : Graphing Absolute Value Functions
Refer to the above figure.
Which of the following functions is graphed?
The correct answer is not given among the other responses.
Below is the graph of :
The given graph is the graph of translated by moving the graph 7 units left (that is, unit right) and 2 units down (that is, units up)
The function graphed is therefore
where . That is,
Example Question #3 : Graphing Absolute Value Functions
What is the equation of the above function?
The formula of an absolute value function is where m is the slope, a is the horizontal shift and b is the vertical shift. The slope can be found with any two adjacent integer points, e.g. and , and plugging them into the slope formula, , yielding . The vertical and horizontal shifts are determined by where the crux of the absolute value function is. In this case, at , and those are your a and b, respectively.
Example Question #4 : Graphing Absolute Value Functions
Give the vertex of the graph of the function .
None of the other choices gives the correct response.
Let
The graph of this basic absolute value function is a "V"-shaped graph with a vertex at the origin, or the point with coordinates . In terms of ,
The graph of this function can be formed by shifting the graph of left 6 units ( ) and down 7 units (). The vertex is therefore located at .