SOLUTION: The parabola given by the equation y=3x^2-30 x+18 has its vertex at (h,k) for: h= and k=

Question 923858: The parabola given by the equation y=3x^2-30 x+18 has its vertex at (h,k) for:
h=
and
k=
Found 2 solutions by ewatrrr, Cromlix:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
the vertex form of a Parabola opening up(a>0) or down(a<0),
where(h,k) is the vertex and x = h is the Line of Symmetry
............
Finding Vertex Form by Completing the Square
y= 3x^2-30x+18
y = 3(x-5)^2 - 75 + 18
y = 3(x-5)^2 -57
V(5, - 57)

Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
y = 3x^2 -30x + 18
h = -b/2a = 30/6 = 5
Substitute x = 5 into 3x^2 -30x + 18
3(5)^2 -30(5) + 18 = -57
So vertex is at (5, -57)
Hope this helps
:-)
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