Question 836304
In general, if you have y = ax^2 + bx + c, the axis of symmetry is x = -b/(2a)


In this case, a = -3, b = -6



x = -b/(2a)


x = -(-6)/(2(-3))


x = 6/(-6)


x = -1


The axis of symmetry, and the x coordinate of the vertex, is x = -1


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Plug this into the given equation to find the y coordinate of the vertex


y = -3x^2-6x-5


y = -3(-1)^2-6(-1)-5


y = -3(1)-6(-1)-5


y = -3 + 6 - 5


y = -2


So the vertex is <font color="red">(-1,-2)</font>