Question

vertex of x^2 + 6x - 16 = 0

Answer 1

The vertex of a parabola is right in the middle of the two x-intercepts. So solve first for those by factoring:
x^2+6x-16 = (x+8)(x-2) = 0

So your intercepts occur at x = -8, and 2. Then the halfway point between -8 and 2 is -3. Therefore, the x-coordinate of the vertex is -3!

Plug in -3 to find the y-coordinate and you're done :)

Answer 2

wow
what is wrong with
\[-\frac{b}{2a}\]?