13 pts! i need with the graphing .

1) Vertex: (0,2)
AOS x=0
table : x = -1, 0, 1
x^2+2 = 3, 2, 3
(x,y) = (-1,3), (0,2), (1,3)
2) vertex: (-5, -1)
AOS x= -5
table : x = -7, -6, -5, -4, -3
x^2+10x+24 = 3, 0, -1, 0, 3
(x,y) = (-7,3), (-6,0), (-5,-1), (4,0), (-3,3)
3) vertex: (4, -1)
AOS x=4
table : x = 2, 3, 4, 5, 6
y = 3, 0, -1, 0, 3
(x,y) = (2, 3), (3, 0), (4, -1), (5, 0), (6, 3)
if the picture didn’t attach along with this then just use the points i gave (x,y) to plot a graph. always remember that whenever your equation has x^2 in it, it is a parabola. and do not forget to write down “y=“. my teacher takes off a lot of points if i don’t. have a great day!

Natlpatlpa is right.

The answer would be the second choice : )

You would need a laid out graph for this but could you provide a picture of just a graph? i can give stepssolve for y.subtract 4xdivide by -1 (opposites)graph slope intercept formcheck each point to see if it’s on the graph