Can someone please tell me what the vertex and line of symentry would be for r(x)=-3(x+3)^2.. Thank you
F(x) = y = -3(x + 3)^2 + k.
Vertex Form:
y = a(x - h)^2 + k.
V(h , k) = V(-3 , 0),
Line of sym = h = -3 = Xv.
h is positive in your Eq because you are subtracting a negative 3.
If you gave me the complete Eq, k = 0.
The graph opens downward since a is negative(-3).