Find the vertex for the parabola whose equation is given y=3x^2+6x-4 A. (-1,-7) B. (1,5) C. (2,20) D. (-2,2)

To find the vertex of the parabola given by the equation $y=3x^{2}+6x-4$, we can use the vertex formula for a quadratic equation in the form $y=ax^{2}+bx+c$. The x-coordinate of the vertex is given by $x=-\frac{b}{2a}$, and the y-coordinate is found by substituting this x-value back into the original equation.<br /><br />Given:<br />$y=3x^{2}+6x-4$<br /><br />Here, $a=3$, $b=6$, and $c=-4$.<br /><br />Step 1: Calculate the x-coordinate of the vertex.<br />$x=-\frac{b}{2a}$<br />$x=-\frac{6}{2(3)}$<br />$x=-\frac{6}{6}$<br />$x=-1$<br /><br />Step 2: Calculate the y-coordinate of the vertex by substituting $x=-1$ back into the original equation.<br />$y=3(-1)^{2}+6(-1)-4$<br />$y=3(1)-6-4$<br />$y=3-6-4$<br />$y=-7$<br /><br />Therefore, the vertex of the parabola is $(-1,-7)$.<br /><br />The correct answer is:<br />A. $(-1,-7)$