How do you find the vertex of #y=3x^2 - 12x - 2#?

Question

Hazel Anglin · Accepted Answer

The vertex is at #(2, -14)#.

To find the vertex of a standard quadratic equation #y = ax^2 + bx + c#, we first use the formula #-b/(2a)# to find the #x# value of the vertex, or #x_v#. We know that #a = 3# and #b = -12#: #x = -b/(2a) = (-(-12))/(2(3)) = 12/6 = 2# To find the #y# value of the vertex we simply plug in the #x# value back into the equation and solve for #y#: #y = 3x^2 - 12x - 2# #y = 3(2)^2 - 12(2) - 2# #y = 3(4) - 24 - 2# #y = 12 - 24 - 2# #y = -14# Therefore, the vertex is at #(2, -14)#. Hope this helps!

Kennedy Adams · Answer

undefined To find the vertex of the quadratic equation $y = 3x^2 - 12x - 2$, you can use the formula for the x-coordinate of the vertex, which is given by $x = -\frac{b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic equation. In this case, $a = 3$ and $b = -12$. After finding $x$, substitute it into the equation to find the corresponding $y$-coordinate. Thus, the vertex of the given quadratic equation is $(2, -20)$.