Absolute Value - Algebra 2

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.

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,

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

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

The equation can be determined from the graph by following the rules of transformations; the base equation is:

The graph of this base equation is:

When we compare our graph to the base equation graph, we see that it has been shifted right 3 units, up 1 unit, and our graph has been stretched vertically by a factor of 2. Following the rules of transformations, the equation for our graph is written as:

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 .

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 ,

,

or, alternatively written,

The graph of is the same as that of , after it shifts 10 units left ( ), it flips vertically (negative symbol), and it shifts up 10 units (the second ). The flip does not affect the position of the vertex, but the shifts do; the vertex of the graph of is at .

The inequality compares an absolute value function with a negative integer. Since the absolute value of any real number is greater than or equal to 0, it can never be less than a negative number. Therefore, can never happen. There is no solution.

Solve for positive values by ignoring the absolute value. Solve for negative values by switching the inequality and adding a negative sign to 7.

First, subtract 5 from both sides to get the absolute value expression alone.

Split this into two linear equations:

or

The solution set is

We start by finding the midpoint of the interval, which is enclosed by 90 and 210. We find the midpoint, or average, of these two endpoints by adding them and dividing by two:

150 is exactly 60 units away from both endpoints, 90 and 210. Since we are looking for the range of numbers that fall in between 90 and 210, this means that any possible value can't be more than 60 units away from 150. If a number is more than 60 units away from 150, in either the increasing or decreasing direction, it will be outside of the \[90, 210\] interval. We can express this using absolute value in the following way:

Divide both sides by 3.

Consider both the negative and positive values for the absolute value term.

Subtract 2 from both sides to solve both scenarios for .

The absolute value gives two problems to solve. Remember to switch the "less than" to "greater than" when comparing the negative term.

or

Solve each inequality separately by adding to all sides.

or

This can be simplified to the format .

Remove the absolute value by setting the term equal to either or . Remember to flip the inequality for the negative term!

Solve each scenario independently by subtracting from both sides.

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

The absolute value of any number is nonnegative, so must always be greater than . Therefore, any value of makes this a true statement.

To solve this inequality, it is best to break it up into two separate inequalities to eliminate the absolute value function:

or .

Then, solve each one separately:

Combining these solutions gives: