Question 116774
Since the quadratic is in vertex form {{{a(x-h)^2+k}}} where (h,k) is the vertex, we can use this general formula to find the vertex of {{{y=-2(x+2)^2 +2}}}


{{{y=-2(x+2)^2 +2}}} Start with the given equation



{{{y=-2(x-(-2))^2 +2}}} Rewrite +2 as -(-2)


Notice how h=-2 and k=2. So the vertex is (-2,2)



Notice if you graph the equation {{{y=-2(x+2)^2 +2}}} you can see that the vertex is (-2,2). So this verifies our answer.



{{{drawing(900,900,-11,9,-9,11,
grid( 1 ),
graph(900,900,-11,9,-9,11, -2(x+2)^2 +2),

circle(-2,2,0.05),
circle(-2,2,0.08)

)}}}