Answer:
The equation of the quadratic in standard form is:
[tex]y=-2x^2-8x-2[/tex]
Step-by-step explanation:
Since they give us the information about where the vertex of the parabola is located, and one extra points where it passes through, we can use the general form of a quadratic in vertex form:
[tex]y-y_v=a\,(x-x_v)^2[/tex]
where [tex](x_v,y_v)[/tex] is the location of the vertex (in our case the point (-2,6).
Therefore the equation above becomes:
[tex]y-y_v=a\,(x-x_v)^2\\y-6=a\,(x-(-2))^2\\y-6=a\,(x+2)^2[/tex]
Now,we can use the fact that the point (-4,-2) is also a point of the graph, to find the value of the parameter [tex]"a"[/tex]:
[tex]y-6=a\,(x+2)^2\\-2-6=a\,(-4+2)^2\\-8=a\,(-2)^2\\-8=a\,*4\\a=-2[/tex]
Then, the equation of the quadratic with such characteristics becomes:
[tex]y-6=-2\,(x+2)^2\\y-6=-2x^2-8x-8\\y=-2x^2-8x-2[/tex]
which is the equation of the quadratic in standard form.
