Question

match each table with its equation.
$y = x^3$
$y = x$
$y = \frac{1}{x}$
$y = |x|$
$y = \sqrt{x}$
$y = x^2$

a. \

$$\begin{tabular}{|c|c|} \\hline input & output \\\\ \\hline -2 & 4 \\\\ \\hline -1 & 1 \\\\ \\hline 0 & 0 \\\\ \\hline 1 & 1 \\\\ \\hline 2 & 4 \\\\ \\hline 3 & 9 \\\\ \\hline \\end{tabular}$$

b. \

$$\begin{tabular}{|c|c|} \\hline input & output \\\\ \\hline 0 & 0 \\\\ \\hline 1 & 1 \\\\ \\hline 4 & 2 \\\\ \\hline 9 & 3 \\\\ \\hline \\end{tabular}$$

c. \

$$\begin{tabular}{|c|c|} \\hline input & output \\\\ \\hline -2 & -2 \\\\ \\hline -1 & -1 \\\\ \\hline 0 & 0 \\\\ \\hline 1 & 1 \\\\ \\hline 2 & 2 \\\\ \\hline 3 & 3 \\\\ \\hline \\end{tabular}$$

d. \

$$\begin{tabular}{|c|c|} \\hline input & output \\\\ \\hline -2 & $-\\frac{1}{2}$ \\\\ \\hline -1 & -1 \\\\ \\hline 0 & undefined \\\\ \\hline 1 & 1 \\\\ \\hline 2 & $\\frac{1}{2}$ \\\\ \\hline 3 & $\\frac{1}{3}$ \\\\ \\hline \\end{tabular}$$

e. \

$$\begin{tabular}{|c|c|} \\hline input & output \\\\ \\hline -2 & -8 \\\\ \\hline -1 & -1 \\\\ \\hline 0 & 0 \\\\ \\hline \\end{tabular}$$

Explanation:

for Table a:

Step1: Analyze the output for each input

For input \( x = -2 \), output is \( 4 \); \( x = -1 \), output \( 1 \); \( x = 0 \), output \( 0 \); \( x = 1 \), output \( 1 \); \( x = 2 \), output \( 4 \); \( x = 3 \), output \( 9 \). Notice that \( (-2)^2 = 4 \), \( (-1)^2 = 1 \), \( 0^2 = 0 \), \( 1^2 = 1 \), \( 2^2 = 4 \), \( 3^2 = 9 \). So the equation is \( y = x^2 \).

for Table b:

Step1: Analyze the output for each input

For input \( x = 0 \), output \( 0 \); \( x = 1 \), output \( 1 \); \( x = 4 \), output \( 2 \); \( x = 9 \), output \( 3 \). Notice that \( \sqrt{0} = 0 \), \( \sqrt{1} = 1 \), \( \sqrt{4} = 2 \), \( \sqrt{9} = 3 \). So the equation is \( y=\sqrt{x} \).

for Table c:

Step1: Analyze the output for each input

For input \( x = -1 \), output \( -1 \); \( x = 0 \), output \( 0 \); \( x = 1 \), output \( 1 \); \( x = 2 \), output \( 2 \); \( x = 3 \), output \( 3 \). Notice that \( y = x \) (since output equals input). So the equation is \( y = x \).

for Table d:

Step1: Analyze the output for each input

For input \( x = -2 \), output \( -\frac{1}{2} \); \( x = -1 \), output \( -1 \); \( x = 0 \), undefined; \( x = 1 \), output \( 1 \); \( x = 2 \), output \( \frac{1}{2} \); \( x = 3 \), output \( \frac{1}{3} \). Notice that \( y=\frac{1}{x} \) (since \( \frac{1}{-2}=-\frac{1}{2} \), \( \frac{1}{-1}=-1 \), \( \frac{1}{1}=1 \), \( \frac{1}{2}=\frac{1}{2} \), \( \frac{1}{3}=\frac{1}{3} \)).

Answer:

• Table a: \( y = x^2 \)
• Table b: \( y=\sqrt{x} \)
• Table c: \( y = x \)
• Table d: \( y=\frac{1}{x} \)
• Table e: \( y = x^3 \) (assuming the remaining inputs follow \( y = x^3 \), e.g., if \( x = 1 \), \( y = 1^3 = 1 \); \( x = 2 \), \( y = 2^3 = 8 \) etc.) Also, for the absolute value function \( y = |x| \), we can check if there's a table left, but from the given tables, the above matches.