Question

find the vertex and the axis of symmetry of the graph of $y = -6(x + 4)^2 - 3$. the vertex is $(square, square)$. the axis of symmetry is $x = square$.

Explanation:

Step1: Recall the vertex form of a parabola

The vertex form of a parabola is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex and the axis of symmetry is \( x = h \).

Step2: Identify \( h \) and \( k \) from the given equation

The given equation is \( y = -6(x + 4)^2 - 3 \). We can rewrite \( (x + 4) \) as \( (x - (-4)) \). So comparing with \( y = a(x - h)^2 + k \), we have \( h = -4 \) and \( k = -3 \).

Step3: Determine the vertex and axis of symmetry

From the vertex form, the vertex is \((h, k)\), so substituting \( h = -4 \) and \( k = -3 \), the vertex is \((-4, -3)\). The axis of symmetry is \( x = h \), so substituting \( h = -4 \), the axis of symmetry is \( x = -4 \).

Answer:

The vertex is \((-4, -3)\).
The axis of symmetry is \( x = -4 \).