Question

Identify the vertex: y = (x+10)2+ 2 a. (10,2) b. (-10, -2) c. (-10,2) ) d. (10, -2)

Answer 1

Given a quadratic equation in vertex form

`y=a(x-h)^2+k`
`\begin{gathered} \text{ wh}ere\text{ a, h, and k are constants, } \\ \text{then the point (h, k) is the vertex of the function} \end{gathered}`

In our case,

`y=(x+10)^2+2`

h = -10, and k= 2

Therefore the vertex is (-10, 2)