To find the quadratic function of the parabola given the vertex [tex]\((-3, 1)\)[/tex] and the directrix [tex]\(y = 6\)[/tex], let's go through the steps to derive the function.
1. Identify the vertex and directrix:
- The vertex is given as [tex]\((-3, 1)\)[/tex].
- The directrix is given as [tex]\(y = 6\)[/tex].
2. Determine the distance [tex]\(p\)[/tex] between the vertex and the directrix:
- [tex]\(p\)[/tex] is the vertical distance from the vertex to the directrix. Since the vertex is at [tex]\(y = 1\)[/tex] and the directrix is at [tex]\(y = 6\)[/tex], we find [tex]\(p\)[/tex] as follows:
[tex]\[
p = 1 - 6 = -5
\][/tex]
This tells us that the focus is 5 units below the vertex (negative distance indicates the direction).
3. Formulate the standard equation of the parabola:
- The standard form of a parabola with vertex [tex]\((h, k)\)[/tex] and distance [tex]\(p\)[/tex] is:
[tex]\[
(x - h)^2 = 4p(y - k)
\][/tex]
- Substituting [tex]\(h = -3\)[/tex], [tex]\(k = 1\)[/tex], and [tex]\(p = -5\)[/tex] into the equation, we get:
[tex]\[
(x + 3)^2 = 4(-5)(y - 1)
\][/tex]
Simplifying, this becomes:
[tex]\[
(x + 3)^2 = -20(y - 1)
\][/tex]
4. Convert to the function form [tex]\(y = f(x)\)[/tex]:
- Solve the equation for [tex]\(y\)[/tex]:
[tex]\[
y - 1 = \frac{1}{-20}(x + 3)^2
\][/tex]
[tex]\[
y = \frac{1}{-20}(x + 3)^2 + 1
\][/tex]
- Simplifying further:
[tex]\[
y = -\frac{1}{20}(x + 3)^2 + 1
\][/tex]
Thus, the quadratic function of the parabola is:
[tex]\[
f(x) = -0.05(x + 3)^2 + 1
\][/tex]
So, the final answer is:
[tex]\[
f(x) = -0.05(x + 3)^2 + 1
\][/tex]
