Question

question 3 4-1 quiz review - graphing linear functions fill in the missing values in the table for $f(x) = 2x - 4$. \

$$\begin{tabular}{|l|l|l|l|l|l|} \\hline $x$ & $-2$ & $0$ & $2$ & $4$ & $6$ \\\\ \\hline $f(x)$ & $-8$ & $-4$ & \\multicolumn{1}{|c|}{} & \\multicolumn{1}{|c|}{} & \\multicolumn{1}{|c|}{} \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Substitute \( x = 2 \) into \( f(x) = 2x - 4 \)

\( f(2)=2\times2 - 4 \)
\( f(2)=4 - 4 = 0 \)

Step2: Substitute \( x = 4 \) into \( f(x) = 2x - 4 \)

\( f(4)=2\times4 - 4 \)
\( f(4)=8 - 4 = 4 \)

Step3: Substitute \( x = 6 \) into \( f(x) = 2x - 4 \)

\( f(6)=2\times6 - 4 \)
\( f(6)=12 - 4 = 8 \)

Answer:

For \( x = 2 \), \( f(x)=0 \); for \( x = 4 \), \( f(x)=4 \); for \( x = 6 \), \( f(x)=8 \)

So the completed table has \( f(x) \) values \( 0 \), \( 4 \), \( 8 \) corresponding to \( x = 2 \), \( x = 4 \), \( x = 6 \) respectively.