Answer:
[tex]y=x^2+6x+8[/tex]
Step-by-step explanation:
To write the quadratic in standard form, begin by writing it in vertex form
[tex]y = a(x-h)^2+k[/tex]
Where (h,k) is the vertex of the parabola.
Here the vertex is (-3,-1). Substitute and write:
[tex]y=a(x--3)^2+-1\\y=a(x+3)^2-1[/tex]
To find a, substitute one point (x,y) from the parabola into the equation and solve for a. Plug in (-2,0) a x-intercept of the parabola.
[tex]0=a((-2)+3)^2-1\\0=a(1)^2-1\\0=a-1\\1=a[/tex]
The vertex form of the equation is [tex]y=(x+3)^2-1[/tex].
To write in standard form, convert vertex form through the distributive property.
[tex]y=(x+3)^2-1\\y=x^2+6x+9-1\\y=x^2+6x+8[/tex]
