a. Given the equation [tex]y = x^2 - 3[/tex], complete the following table:

\begin{tabular}{|l|l|l|l|l|l|}
\hline
[tex]$x$[/tex] & -5 & -3 & 0 & 1 & 2 \\
\hline
[tex]$y$[/tex] & 22 & 6 & -3 & -2 & 1 \\
\hline
\end{tabular}



Answer :

Let's determine the missing [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values using the function [tex]\( y = x^2 - 3 \)[/tex]. We need to fill in the entries of the table.

Given:
[tex]\[ y = x^2 - 3 \][/tex]

We are provided with:
[tex]\[\begin{array}{|c|c|c|c|c|c|} \hline x & -5 & -3 & 0 & 1 & 2 \\ \hline y & \text{?} & \text{?} & \text{?} & 1 & 13 \\ \hline \end{array}\][/tex]

### Step-by-Step Solution:

1. For [tex]\( x = -5 \)[/tex]:
[tex]\[ y = (-5)^2 - 3 \][/tex]
[tex]\[ y = 25 - 3 \][/tex]
[tex]\[ y = 22 \][/tex]

Therefore:
[tex]\[ y = 22 \][/tex]

2. For [tex]\( x = -3 \)[/tex]:
[tex]\[ y = (-3)^2 - 3 \][/tex]
[tex]\[ y = 9 - 3 \][/tex]
[tex]\[ y = 6 \][/tex]

Therefore:
[tex]\[ y = 6 \][/tex]

3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = (0)^2 - 3 \][/tex]
[tex]\[ y = 0 - 3 \][/tex]
[tex]\[ y = -3 \][/tex]

Therefore:
[tex]\[ y = -3 \][/tex]

### Final Table:

[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -5 & -3 & 0 & 1 & 2 \\ \hline y & 22 & 6 & -3 & 1 & 13 \\ \hline \end{array} \][/tex]

Thus, the completed table is:

[tex]\[ \begin{array}{| c | c | c | c | c | c |} \hline x & -5 & -3 & 0 & 1 & 2 \\ \hline y & 22 & 6 & -3 & 1 & 13 \\ \hline \end{array} \][/tex]