Mathematical Relationships and Basic Graphs - Algebra 2

Multiply the first number with the ones digit of the second number.

Repeat the process with the tens digit.

Add a zero to the end of this number and add it with the first number.

The answer is:

Multiply the first number with the ones digit of the second number.

Repeat the process for the tens digit of the second number.

Add a zero to this number and add the value with the first number.

The answer is:

Multiply the first number with the ones digit of the second number.

Repeat the process for the tens digit.

Add an extra zero to the end of this number.

Add this value with the first number.

The answer is:

Multiply the first number with the ones digit of the second number.

Repeat the process for the tens digit of the second number.

Add a zero to the end of this number and add the first number.

The answer is:

Multiply the first number with the ones digit of the second number.

Save this value.

Multiply the first number with the tens digit of the second number.

Add a zero to this number and save the value for later.

Multiply the first number with the hundreds digit of the second number.

Add two zeros to the end of this number, and add the previously saved numbers.

The answer is:

Multiply the first number given with the ones digit of 73.

Multiply the first number given with the tens digit of 73.

Add an extra zero to the end of this number and add this with the first number.

The answer is:

The quotient of any number and zero is considered an undefined quantity.

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

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 .