Find the vertex of the parabola y=-6x^2+2 Simplify both coordinates and write them as proper fractions, improper fractions, o square square

To find the vertex of the parabola given by the equation $y = -6x^2 + 2$, we can use the vertex formula for a quadratic equation in the form $y = ax^2 + bx + c$.<br /><br />The vertex formula is given by:<br />$x_v = -\frac{b}{2a}$<br />$y_v = c - \frac{b^2}{4a}$<br /><br />In this case, the equation is $y = -6x^2 + 2$, which is in the form $y = ax^2 + bx + c$ with $a = -6$, $b = 0$, and $c = 2$.<br /><br />Substituting these values into the vertex formula, we get:<br />$x_v = -\frac{0}{2(-6)} = 0$<br />$y_v = 2 - \frac{0^2}{4(-6)} = 2$<br /><br />Therefore, the vertex of the parabola is $(0, 2)$.