Answer:
[tex]y = (-2)(x + 2)^2 - 4[/tex].
Step-by-step explanation:
The vertex form of a quadratic function is in the form
[tex]y = a (x - h)^2 + k[/tex],
where
In this question, the vertex of this quadratic function is at the point [tex](-2, -4)[/tex]. In other words, [tex]h = (-2)[/tex] and [tex]k = (-4)[/tex]. Substitute these value into the general equation:
[tex]y = a (x - (-2))^2 +(- 4)[/tex].
Simplify to obtain:
[tex]y = a (x + 2)^2 - 4[/tex].
The only missing piece here is the coefficient [tex]a[/tex]. That's likely why the problem gave [tex](-1, -6)[/tex], yet another point on this quadratic function. If this function indeed contains the point [tex](-1, -6)[/tex], [tex]y[/tex] should be equal to [tex](-6)[/tex] when [tex]x = (-1)[/tex]. That is:
[tex]-6 = a(-1 + 2)^2 -4[/tex].
Solve this equation for [tex]a[/tex]:
[tex]a = -6 - (-4) = -2[/tex].
Hence the equation of the quadratic function in its vertex form:
[tex]y = (-2)(x + 2)^2 - 4[/tex].
