Question

Complete each table by evaluating the expression using the given values for the variable. $$\begin{array}{|r|r|} \hline t & |2 t-4| \\ \hline-2 & \\ \hline 0 & \\ \hline 2 & \\ \hline 4 & \\ \hline 6 & \\ \hline \end{array}$$

Short answer

\\ \hline 0 & ? \\ \hline 2 & ? \\ \hline 4 & ? \\ \hline 6 & ? \\ \hline \end{array}$$ Answer: $$\begin{array}{|r|r|} \hline t & |2t-4| \\ \hline-2 & 8 \\ \hline 0 & 4 \\ \hline 2 & 0 \\ \hline 4 & 4 \\ \hline 6 & 8 \\ \hline \end{array}$$

Step-by-step solution

Understand the expression

The given expression is $$|2t - 4|$$, which represents the absolute value of the linear function $$2t - 4$$. To evaluate this expression, we will substitute the given values of $$t$$ in the table and find the absolute value of the result.

Evaluate the expression for given t values

Now we will substitute the given values of $$t$$ and find the corresponding values of the expression. When $$t=-2$$: $$|2(-2) - 4| = |-4 - 4| = |-8| = 8$$ When $$t=0$$: $$|2(0) - 4| = |0 - 4| = |-4| = 4$$ When $$t=2$$: $$|2(2) - 4| = |4 - 4| = |0| = 0$$ When $$t=4$$: $$|2(4) - 4| = |8 - 4| = |4| = 4$$ When $$t=6$$: $$|2(6) - 4| = |12 - 4| = |8| = 8$$

Fill the complete table

Now we have computed the values for the given expression, we can fill the table: $$\begin{array}{|r|r|} \hline t & |2 t-4| \\ \hline-2 & 8 \\ \hline 0 & 4 \\ \hline 2 & 0 \\ \hline 4 & 4 \\ \hline 6 & 8 \\ \hline \end{array}$$

Key concepts

Understanding Absolute Value

The absolute value of a number refers to its distance from zero on the number line, without considering direction. This means:

• It's always a positive number or zero.
• Denoted by vertical bars, such as \(|x|\), which means the absolute value of \(x\).

For example, the absolute value of -3 is 3 because -3 is 3 units away from zero. Likewise, the absolute value of 3 is also 3.
When evaluating expressions like \(|2t - 4|\), we first calculate \(2t - 4\) and then take the absolute value.

• If the result is positive or zero, it stays the same.
• If it's negative, we drop the negative sign.

This simple concept helps in transforming any result into its non-negative counterpart, making it crucial when evaluating expressions.

Mastering Variable Substitution

Variable substitution is a key concept in algebra, used to simplify expressions by replacing variables with given values.

• Each variable stands for a number that you substitute into the expression.
• This substitution is essential for evaluating the expression accurately.

For the exercise, we use various values of \(t\), like -2, 0, 2, 4, and 6. Substituting these into the expression \(2t - 4\) gives us a defined numeric expression to work with.
Let's see how it works:

• For \(t = -2\), replace \(t\) with -2 to get \(2(-2) - 4\).
• Simplifying it gives \(-8\), and taking the absolute value results in \(8\).

By following these steps, you systematically break down the expression to make calculations straightforward.

Exploring Linear Functions

Linear functions are functions whose graph is a straight line. They have the general form \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
In our exercise, the linear function in question is \(2t - 4\).

• Here, \(2\) is the coefficient or slope that determines the steepness of the line.
• The \(-4\) is the y-intercept, indicating where the line crosses the y-axis.

Linear functions are foundational in algebra, as they describe relationships with constant rates of change.
When we evaluate linear functions, we're finding specific values based on input variables, like \(t\) in \(2t - 4\).
This simple understanding allows us to manipulate and graph these functions confidently, showing their wide-ranging applications in math.