what is the vertex and axis of symmerty of the following quadratic function: f(x) = 3(x+1)^2 - 5
The given quadratic function is in the form f(x) = a(x-h)^2 + k, where a = 3, h = -1, and k = -5.
The vertex form of a quadratic function is f(x) = a(x-h)^2 + k, where (h, k) represents the vertex of the parabola.
Therefore, the vertex is at (-1, -5).
The axis of symmetry is given by the equation x = h.
Therefore, the axis of symmetry is x = -1.
real easy:
at symmetry x+1 = -(x+1) so where 2 x = -2
or x = -1
the vertex is therefore at x = -1
so
y = 3(-1+1)^2 - 5 = -5
so at ( -1 , -5 )