Questions: Given the function (y=-frac14(x+3)^2-5), what is the vertex, focus, axis of symmetry and directrix?

Solution Steps

To solve this problem, we need to identify the vertex, focus, axis of symmetry, and directrix of the given quadratic function \( y = -\frac{1}{4}(x+3)^2 - 5 \). The vertex form of a parabola is \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex. The focus and directrix can be found using the properties of parabolas.

1. Identify the vertex \((h, k)\) from the given equation.
2. Determine the value of \(a\) to find the distance from the vertex to the focus and directrix.
3. Calculate the coordinates of the focus.
4. Determine the equation of the directrix.
5. Identify the axis of symmetry.

The given function is \( y = -\frac{1}{4}(x+3)^2 - 5 \). This is in the form \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex. Here, \( h = -3 \) and \( k = -5 \).

Thus, the vertex is: \[ \text{Vertex} = (-3, -5) \]

The distance from the vertex to the focus and the directrix is given by: \[ \text{distance} = \frac{1}{4|a|} \] Given \( a = -\frac{1}{4} \), we have: \[ \text{distance} = \frac{1}{4 \times \frac{1}{4}} = 1 \]

The focus of a parabola \( y = a(x-h)^2 + k \) is located at \( (h, k + \frac{1}{4a}) \) if \( a > 0 \) and \( (h, k - \frac{1}{4a}) \) if \( a < 0 \). Since \( a = -\frac{1}{4} \), the focus is: \[ \text{Focus} = (h, k - \text{distance}) = (-3, -5 - 1) = (-3, -6) \]

The directrix of the parabola is given by: \[ y = k + \text{distance} \] Thus, the directrix is: \[ \text{Directrix} = y = -5 + 1 = -4 \]

The axis of symmetry for the parabola \( y = a(x-h)^2 + k \) is the vertical line \( x = h \). Therefore, the axis of symmetry is: \[ \text{Axis of Symmetry} = x = -3 \]

Final Answer