find the vertex y+-5(x+2)^2+5

Question

Identifying the vertex and the y-intercept.

Answer 1

I assume the equation y  = -5(x + 2)^2  + 5 ---> (1)

y  = -5(x -(-2))^2+5

Compare it to parabola equation y  = a (x - h)^2 + k.

Vertex of parabola ( h , k ) = (-2,5)

To find y  intercept substitute x  = 0 in (1).

y  = -5(0+2)^2 +5

y  = -5(4) +5

y  = -20+5

y  = -15

y  intercept is -15.