Question

What is the vertex for the graph of y– 4 = - (x+1)^2??O A. (4,-1)O B. (1,-4)O c. (-1,4)O D. (-4,1)

Answer 1

The equiation of parabola in vertex form is:

`y\text{ = a}\cdot(x-h)^2\text{ + k}`

where (h, k) are the coordinates of the vertex and a is a constant. In our case we have:

`y-4=-(x+1)^2`

add 4 to both sides:

`y\text{ -4 +4 = -}(x+1)^2\text{ + 4}`

Resolving:

`y=-(x+1)^2\text{ + 4}`

Therefore we already have it in the vertex form and we have to:

`\begin{gathered} h\text{ = -1 } \\ k\text{ = 4} \end{gathered}`

Therefore the answer is c. (-1, 4)