In the equation y=(x+4)^2-3. what is the vertex, zeros, and y intercepts. explain how you got your answer .

For
y=a(x-h)^2+k
vertex=(h,k)

y=(x-(-4))^2+(-3)
vertex is (-4,-3)

xintercept is where y=0
0=(x+4)^2-3
add 3
3=(x+4)^2
sqrt both sides don't forget positive and negative
+/-√3=x+4
minus 4
-4+/-√3=x
x=-4+√3 and x=-4-√3
xints are (-4+√3,0) and (-4-√3,0)

yint is where x=0
y=(0+4)^2-3
y=(4)^2-3
y=16-3
y=13
yint is 13

vertex is (-4,-3)
xints are (-4+√3,0) and (-4-√3,0)
yint is at (0,13)