What is the vertex of -3x^2 - x + 12

Hey there!

To start, first find your axis of symmetry, or the x value of your vertex, which is found by using the formula
x=-b/2a.

In the quadratic equation -3x^2-x+12, the b value would be -1 (the b value is the second coefficient as according to ax^2+bx+c) and the a value would be -3.

Now, plug the values into the axis of symmetry formula and simplify for x:
x=-(-1)/2(-3)
x=1/-6
x=-

Now, plug the value of the axis of symmetry to solve for the y value of the vertex:
-3(-1/6)^2-(-1/6)+12
-3(1/36)+1/6+12
-1/12+1/6+12
1/12+12
=

Knowing that the x value, or axis of symmetry is - and the y value is , your vertex would be (-,).

Hope this helps and I hope you have a wonderful day! :)