Answer:
[tex]\large\boxed{y=(x+3)^2=x^2+6x+9}[/tex]
Step-by-step explanation:
It's the parabola - the graph of a quadratic function.
The vertex form of an equation of a parabola:
[tex]y=a(x-h)^2+k[/tex]
(h, k) - vertex
We have the vertex in (-3, 0) → h = -3, k = 0. Substitute:
[tex]y=a(x-(-3))^2+0\\\\y=a(x+3)^2[/tex]
Choice one point from the graph (-1, 4). Substitute:
[tex]4=a(-1+3)^2\\\\4=a(2)^2[/tex]
[tex]4=a(4)[/tex] divide both sides by 4
[tex]1=a\to a=1[/tex]
Finally:
[tex]y=(x+3)^2=x^2+6x+9[/tex]
[tex]_{\text{used}\ (a+b)^2=a^2+2ab+b^2}[/tex]
