Find the coordinates of the vertex of the following parabola algebraically. Write your answer as an (x,y) y=x^2-6 Answer Attemptiout of 2 square

To find the coordinates of the vertex of the parabola given by the equation $$y = x^2 - 6$$ , we can use the vertex formula for a parabola in the form $$y = ax^2 + bx + c$$ . The vertex formula is given by:

$$x = -\frac{b}{2a}$$

For the given equation $$y = x^2 - 6$$ , we have $$a = 1$$ , $$b = 0$$ , and $$c = -6$$ . Plugging these values into formula, we get:

$$x = -\frac{0}{2 \cdot 1} = 0$$

Now that we have the of the vertex, we can find the y-coordinate by substituting $$x = 0$$ back into the original equation:

$$y = (0)^2 - 6 = -6$$

Therefore, the coordinates of the vertex of the parabola $$y = x^2 - 6$$ are $$(0, -6)$$ .