Question

y-k=a(x-h)2,a=1,h<0,k>0, in which quardrant is the vertex?

Answer 1

This might help; http://openstudy.com/study#/updates/4f6aa0cee4b014cf77c7d01f

Answer 2

rewriting the equation into vertex form you get

\[y = a(x - h)^2 + k\]

the vertex is at (h, k)

is h < 0 this is the x value... so it it left of the origin

k > 0 this is the y have in the vertex so this is above the origin

so you need to locate a point left of the origin and above the origin

so which quadrant is that..?