What is the vertex of y= 1/3(x/5+1)^2+4/15 ?

1 Answer
Jan 2, 2016

The vertex form is y = a(x-h)^2 +k where (h,k) is the vertex. For our problem the vertex is (-5,4/15)

Explanation:

y=1/3(x/5+1)^2 + 4/15

y=1/3((x+5)/5)^2 + 4/15

y=1/75(x+5)^2 + 4/15

Compare with
y=a(x-h)^2 + k

h=-5 and k=4/15

The vertex (h,k) is (-5,4/15)