Question

Y = 3x² + 6x + 4 has a O maximum O minimum at

Answer 1

In the equation:

y = 3x² + 6x + 4

the leading coefficient, a, is equal to 3. Given that a is greater than zero, then the parabola has a shape of a U. Therefore, the parabola has a minimum.

To find the minimum, we need to find the vertex (h, k).

The x-coordinate of the vertex, h, is found as follows:

`\begin{gathered} h=(-b)/(2a) \\ h=(-6)/(2\cdot3) \\ h=-1 \end{gathered}`

The y-coordinate of the vertex, k, is found substituting h into the equation of the parabola, as follows:

`\begin{gathered} y=3x^2+6x+4 \\ k=3h^2+6h+4 \\ k=3\cdot(-1)^2+6\cdot(-1)+4 \\ k=3\cdot1-6+4 \\ k=1 \end{gathered}`

The minimum is placed at (-1, 1)