Find the vertex and axis of symmetry and determine the direction that the parabola opens.
$y={(x+4)}^2-9$

• A. Parabola opens downwards; Vertex: $\left(-4,-9\right)$; Axis of symmetry: $x=-4$
• B. Parabola opens upwards; Vertex: $(-4,-9)$; Axis of symmetry: $x=-4$
• C. Parabola opens upwards; Vertex: $(4,-9)$; Axis of symmetry: $x=4$
• D. Parabola opens downwards; Vertex: $\left(4,-9\right)$; Axis of symmetry: $x=4$

Answer: (B)

Verified step by step guidance

1. Identify the given quadratic function in vertex form: \(y = \left(x + 4\right)^2 - 9\).
2. Recall that the vertex form of a parabola is \(y = a\left(x - h\right)^2 + k\), where \((h, k)\) is the vertex.
3. Rewrite the expression inside the parentheses to match the form \(x - h\): since it is \(x + 4\), this means \(h = -4\).
4. The vertex is therefore at the point \((h, k) = (-4, -9)\).
5. Determine the direction the parabola opens by looking at the coefficient \(a\) in front of the squared term. Here, \(a = 1\) (positive), so the parabola opens upwards. The axis of symmetry is the vertical line \(x = -4\).