Solution
Step 1:
Draw and complete the chart
Explanation
[tex]\begin{gathered} \text{For x = -2, x}^2\text{ = \lparen-2\rparen}^2\text{ = 4 , 2x}^2\text{ = 2}\times4\text{ = 8} \\ For\text{ x = -1 , x}^2\text{ = \lparen-1\rparen}^2\text{ = 1, 2x}^2\text{ = 2}\times\text{ 1 = 1} \\ For\text{ x = 0, x}^2\text{ = 0}^2\text{ = 0 , 2x}^2\text{ = 2}\times\text{ 0 = 0} \\ For\text{ x = 1, x}^2\text{ = 1}^2\text{ = 1, 2x}^2\text{ = 2 }\times\text{ 1 = 2} \\ For\text{ x = 2 , x}^2\text{ = 2}^2\text{ = 4, 2x}^2\text{ = 2}\times\text{ 4 = 8} \end{gathered}[/tex][tex]\begin{gathered} \text{y = 3x}^2+\text{ 12x - 4} \\ To\text{ find the minimum value, find the first derivative} \\ \frac{dy}{dx}\text{ = 6x + 12} \\ 6x\text{ + 12 = 0} \\ 6x\text{ = -12} \\ x\text{ = }\frac{-12}{6}\text{ = -2} \end{gathered}[/tex]
Substitute x = -2, to find the minimum value.
[tex]\begin{gathered} \text{y = 3x}^2+12x-4 \\ \text{y = 3}\times(-2)^2\text{ + 12}\times(-2)\text{ - 4} \\ \text{y = 12 - 24 - 4} \\ \text{y = -16} \end{gathered}[/tex]
The minimum value is (-2, -16)
