Write a quadratic function from its vertex and another point

Answer:

$y = (-2)(x + 2)^2 - 4$ .

Step-by-step explanation:

The vertex form of a quadratic function is in the form

$y = a (x - h)^2 + k$ ,

where

$a$ is a coefficient that needs to be found, and
$(h, k)$ is the vertex of this function.

In this question, the vertex of this quadratic function is at the point $(-2, -4)$ . In other words, $h = (-2)$ and $k = (-4)$ . Substitute these value into the general equation:

$y = a (x - (-2))^2 +(- 4)$ .

Simplify to obtain:

$y = a (x + 2)^2 - 4$ .

The only missing piece here is the coefficient $a$ . That’s likely why the problem gave $(-1, -6)$ , yet another point on this quadratic function. If this function indeed contains the point $(-1, -6)$ , $y$ should be equal to $(-6)$ when $x = (-1)$ . That is:

$-6 = a(-1 + 2)^2 -4$ .

Solve this equation for $a$ :

$a = -6 - (-4) = -2$ .

Hence the equation of the quadratic function in its vertex form:

$y = (-2)(x + 2)^2 - 4$ .

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