Question

complete table of values( x -3 -2 -1 0 1 2 3)
y=x^2 +x-4

Answer 1

FUNCTION

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Solve:

$\implies \sf \large y = {x}^{2} + x - 4$

$\large \begin{align}& \begin{array}{cccccccc} \sf x&-3&-2&-1&0& \\ \sf y & \orange{2}& \orange{ - 2} & \orange{ - 4} & \orange{ - 4} & \end{array} \\ &\begin{array}{cccccccc} \sf x & 1&2&3 \\ \sf y & \orange{ - 2} & \orange{0} & \orange{8} \end{array} \end{align}$

$\:$

When x = -3

$\implies \sf \large y = { - 3}^{2} + ( - 3)- 4$

$\implies \sf \large y = 9 - 3 - 4$

$\implies \sf \large \orange{y = 2}$

$\:$

When x = -2

$\implies \sf \large y = { - 2}^{2} + ( - 2)- 4$

$\implies \sf \large y = 4 - 2- 4$

$\implies \sf \large \orange{y = - 2}$

$\:$

When x = -1

$\implies \sf \large y = { - 1}^{2} + ( - 1)- 4$

$\implies \sf \large y = 1 - 1- 4$

$\implies \sf \large \orange{y = - 4}$

$\:$

When x = 0

$\implies \sf \large y = {0}^{2} + 0- 4$

$\implies \sf \large y = {0} + 0- 4$

$\implies \sf \large \orange{ y = - 4}$

$\:$

When x = 1

$\implies \sf \large y = {1}^{2} + 1- 4$

$\implies \sf \large y = 1 + 1- 4$

$\implies \sf \large \orange{y = - 2}$

$\:$

When x = 2

$\implies \sf \large y = {2}^{2} + 2- 4$

$\implies \sf \large y = 4 + 0- 4$

$\implies \sf \large \orange{y = {0}}$

$\:$

When x = 3

$\implies \sf \large y = {3}^{2} + 3- 4$

$\implies \sf \large y = 9 + 3- 4$

$\implies \sf \large \orange{y = 8}$

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