Answer:
[tex]y=x^2+6x+8[/tex]
Step-by-step explanation:
Notice that this is the graph of a parabola with vertex at the point (-3,-1), and which crosses the x-axis at (-4,0) and at (-2,0)
So we can start by using the vertex form of a parabola with vertex at a general point [tex](x_0,y_0)[/tex]:
[tex]y-y_0=a\,(x-x_0)^2[/tex]
For our case:
[tex]y-y_0=a\,(x-x_0)^2\\y-(-1)=a\,(x-(-3))^2\\y+1=a\,(x+3)^2[/tex]
Now,we can use the fact that the curve passes through (-2,0) as well, in order to find "a" and complete the equation:
[tex]y+1=a\,(x+3)^2\\0+1=a\,(-2+3)^2\\1=a\,(1)^2\\a=1[/tex]
Therefore the equation for the curve is:
[tex]y+1=(x+3)^2\\y=(x+3)^2-1\\y=x^2+6x+9-1\\y=x^2+6x+8[/tex]
which is now written in standard form
