For the quadratic function $$ f(x)=\frac{1}{4} x^{2}+2 $$, what is the focal length $$ c^{\prime} $$ ? What is the focal point? What is the equation of the directrix?

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To analyze the quadratic function $$ f(x) = \frac{1}{4} x^2 + 2 $$, we can rewrite it in the standard form of a parabola. The standard form of a parabola that opens upwards is given by:

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

where $$(h, k)$$ is the vertex of the parabola, and $$a$$ determines the width and direction of the parabola.

In our case, we can see that:

- $$a = \frac{1}{4}$$
- The vertex $$(h, k)$$ is at $$(0, 2)$$ since there are no $$x$$ or constant terms that shift the vertex.

### Focal Length $$c'$$

The focal length $$c'$$ is related to the coefficient $$a$$ by the formula:

$$
c' = \frac{1}{4a}
$$

Substituting $$a = \frac{1}{4}$$:

$$
c' = \frac{1}{4 \cdot \frac{1}{4}} = \frac{1}{1} = 1
$$

### Focal Point

The focal point of the parabola is located at a distance $$c'$$ above the vertex for a parabola that opens upwards. Since the vertex is at $$(0, 2)$$ and $$c' = 1$$, the focal point is:

$$
(0, 2 + 1) = (0, 3)
$$

### Equation of the Directrix

The directrix of a parabola that opens upwards is a horizontal line located $$c'$$ units below the vertex. Therefore, the equation of the directrix is:

$$
y = k - c' = 2 - 1 = 1
$$

### Summary

- Focal length $$c' = 1$$
- Focal point: $$(0, 3)$$
- Equation of the directrix: $$y = 1$$
Focal length $$c' = 1$$, focal point $$(0, 3)$$, directrix $$y = 1$$

For the quadratic function , what is the focal length ? What is the focal
point? What is the equation of the directrix?

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For the quadratic function , what is the focal length ? What is the focal point? What is the equation of the directrix?