Question

What is the the vertex of `y=-(x+2)^2+3x+5`?

Answer 1

Vertex is at `(-0.5,1.25)`

Explanation:

`y=-(x+2)^2+3x+5 or y=-(x^2+4x+4)+3x+5`

or `y=-x^2-4x-4+3x+5 or y=-x^2-x+1` or

`y=-(x^2+x)+1 or y=-(x^2+x+0.5^2)+0.5^2+1` or

`y=-(x+0.5)^2+1.25` . Comparing with vertex form of

equation `f(x) = a(x-h)^2+k ; (h,k)` being vertex we find

here `h=-0.5 , k=1.25 :.` Vertex is at `(-0.5,1.25)`

graph{-(x+2)^2+3x+5 [-10, 10, -5, 5]}