To find the quadratic function of the parabola with the given vertex [tex]\((-3, 1)\)[/tex] and directrix [tex]\(y = 6\)[/tex], let's follow these steps:
1. Determine the distance from the vertex to the directrix:
- The vertex is at [tex]\((h, k) = (-3, 1)\)[/tex].
- The directrix is a horizontal line at [tex]\(y = 6\)[/tex].
- The distance from the vertex to the directrix is [tex]\(|1 - 6|\)[/tex], which is 5 units.
2. Find the value of [tex]\(a\)[/tex]:
- For a parabola with a vertex [tex]\((h, k)\)[/tex], the formula connecting the distance [tex]\(d\)[/tex] from the vertex to the directrix to the coefficient [tex]\(a\)[/tex] is given by:
[tex]\[ \frac{1}{4a} = d \][/tex]
- Here, [tex]\(d = 5\)[/tex], so:
[tex]\[ 4a \cdot 5 = 1 \][/tex]
[tex]\[ 20a = 1 \][/tex]
[tex]\[ a = \frac{1}{20} = 0.05 \][/tex]
3. Write the quadratic function in vertex form:
- The vertex form of a parabolic equation is given by:
[tex]\[ y = a(x - h)^2 + k \][/tex]
- Substituting the vertex [tex]\((h, k) = (-3, 1)\)[/tex] and the calculated value of [tex]\(a = 0.05\)[/tex]:
[tex]\[ y = 0.05(x - (-3))^2 + 1 \][/tex]
[tex]\[ y = 0.05(x + 3)^2 + 1 \][/tex]
Therefore, the quadratic function of the parabola is:
[tex]\[ f(x) = 0.05(x + 3)^2 + 1 \][/tex]
