What is the vertex form of the graphed equation? (1 point) Responses y=3(x+2)^2 −3, y is equal to 3 times open paren x plus 2 close paren squared minus 3, y=3(x−2)^2 −3, y is equal to 3 times open paren x minus 2 close paren squared minus 3, y=−3(x+2)^2 −3 y is equal to negative 3 times open paren x plus 2 close paren squared minus 3 y=−3(x−2)^2 −3

Question

vertex is -2, -3, and opens upwards

Answer 1

Given that the vertex of the parabola is at (-2, -3) and it opens upwards, we know that the coefficient of the squared term in the vertex form equation \(y = a(x - h)^2 + k\) must be positive.

Since the vertex is \((-2, -3)\), this means that \(h = -2\) and \(k = -3\). Therefore, we can write the vertex form as:

\[ y = a(x + 2)^2 - 3 \]

Since we know the parabola opens upwards, the coefficient \(a\) must be positive. The problem states one option that fits this description:

Response: \( y = 3(x + 2)^2 - 3 \)

This matches the vertex form with the vertex at (-2, -3) and opens upwards with \(a = 3\).

Thus, the correct answer is:

y = 3(x + 2)^2 - 3.