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=-[tex] \frac{1}{6} [/tex]
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
=[tex] \frac{145}{12} [/tex]
Knowing that the x value, or axis of symmetry is -[tex] \frac{1}{6} [/tex] and the y value is [tex] \frac{145}{12} [/tex], your vertex would be (-[tex] \frac{1}{6} [/tex],[tex] \frac{145}{12} [/tex]).
Hope this helps and I hope you have a wonderful day! :)
