Answer :
To complete the table of values for the equation [tex]\( y = 3x^2 + 2 \)[/tex], we need to determine the values of [tex]\( y \)[/tex] at specific [tex]\( x \)[/tex] values.
Given the table:
[tex]\[ \begin{array}{c||c|c|c|c|c} x & -3 & -2 & -1 & 0 & 1 \\ \hline y & 29 & A & 5 & B & 5 \\ \end{array} \][/tex]
we need to find the values that replace [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 3(-2)^2 + 2 \][/tex]
Calculating:
[tex]\[ y = 3 \times 4 + 2 = 12 + 2 = 14 \][/tex]
So, [tex]\( A = 14 \)[/tex].
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 3(0)^2 + 2 \][/tex]
Calculating:
[tex]\[ y = 3 \times 0 + 2 = 0 + 2 = 2 \][/tex]
So, [tex]\( B = 2 \)[/tex].
Therefore, the completed table is:
[tex]\[ \begin{array}{c||c|c|c|c|c} x & -3 & -2 & -1 & 0 & 1 \\ \hline y & 29 & \mathbf{14} & 5 & \mathbf{2} & 5 \\ \end{array} \][/tex]
Given the table:
[tex]\[ \begin{array}{c||c|c|c|c|c} x & -3 & -2 & -1 & 0 & 1 \\ \hline y & 29 & A & 5 & B & 5 \\ \end{array} \][/tex]
we need to find the values that replace [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 3(-2)^2 + 2 \][/tex]
Calculating:
[tex]\[ y = 3 \times 4 + 2 = 12 + 2 = 14 \][/tex]
So, [tex]\( A = 14 \)[/tex].
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 3(0)^2 + 2 \][/tex]
Calculating:
[tex]\[ y = 3 \times 0 + 2 = 0 + 2 = 2 \][/tex]
So, [tex]\( B = 2 \)[/tex].
Therefore, the completed table is:
[tex]\[ \begin{array}{c||c|c|c|c|c} x & -3 & -2 & -1 & 0 & 1 \\ \hline y & 29 & \mathbf{14} & 5 & \mathbf{2} & 5 \\ \end{array} \][/tex]
