what would be my ordered pairs when graphing a parabola of y=(x+3)^2+1

Question

shamil98 · Answer

y = a(x-h)^2 + k
where (h,k) = vertex

you have
y = 1(x-(-3))^2 + 1
so
(-3,1) is your vertex

anonymous · Answer

ok thank you and I have tried for x: -1, -2, -3, -4, -5 and if I were to continue to -6 the ordered pair is too high for my graph, also if I try 0 the ordered pairs is again to large for my graph.

shamil98 · Answer

when x = 0
y = (0+3)^2 + 1
y = 10
(0,10)
just make your graph represent larger numbers per dash or something

anonymous · Answer

the graph is automatically designed appropriately for the answers of its questions therefore I am unable to go higher than its limits and it can not be modified. my graph goes from y -axis: 9 to -9 and x-axis: 9 to -9

shamil98 · Answer

not much i can do then.

anonymous · Answer

ok thank you anyways

luigi0210 · Answer

Is this regular algebra or is this conics?