Select the correct answer.

The vertex of a parabola is at the point [tex]\((3, 1)\)[/tex], and its focus is at [tex]\((3, 5)\)[/tex]. What function does the graph represent?

A. [tex]\(f(x) = \frac{1}{16}(x-3)^2 - 1\)[/tex]

B. [tex]\(f(x) = \frac{1}{4}(x+3)^2 - 1\)[/tex]

C. [tex]\(f(x) = \frac{1}{4}(x-3)^2 - 1\)[/tex]

D. [tex]\(f(x) = \frac{1}{16}(x-3)^2 + 1\)[/tex]



Answer :

To determine the correct function that represents the graph of the given parabola, we need to follow a series of steps.

1. Understanding the Vertex Form of a Parabola:
Since the vertex of the parabola is given at [tex]\((3, 1)\)[/tex] and the focus is given at [tex]\((3, 5)\)[/tex], this parabola opens vertically. The general form of a vertically oriented parabolic function given the vertex [tex]\((h, k)\)[/tex] and the parameter [tex]\(p\)[/tex] (distance from the vertex to the focus) is:
[tex]$(x - h)^2 = 4p(y - k)$[/tex]

2. Identifying the Parameters:
- The vertex [tex]\((h, k) = (3, 1)\)[/tex]
- The focus [tex]\((3, 5)\)[/tex]
- Since the focus is at [tex]\((3, 5)\)[/tex] and the vertex is at [tex]\((3, 1)\)[/tex], the distance [tex]\(p\)[/tex] is calculated as:
[tex]$p = 5 - 1 = 4$[/tex]

3. Substituting Parameters into the Vertex Form:
Using the vertex form [tex]\((x - h)^2 = 4p(y - k)\)[/tex]:
[tex]$ (x - 3)^2 = 4 \cdot 4 (y - 1) $[/tex]
[tex]$ (x - 3)^2 = 16(y - 1) $[/tex]

4. Solving for [tex]\(y\)[/tex] to Get the Function [tex]\(f(x)\)[/tex]:
Isolate [tex]\(y\)[/tex] to express it in function form:
[tex]$ y - 1 = \frac{(x - 3)^2}{16} $[/tex]
Therefore:
[tex]$ y = \frac{(x - 3)^2}{16} + 1 $[/tex]

5. Converting to Function Form:
Therefore, the function in [tex]\(f(x)\)[/tex] form is:
[tex]$ f(x) = \frac{1}{16}(x - 3)^2 + 1 $[/tex]

Thus, the correct answer is:
D. [tex]\( f(x) = \frac{1}{16}(x - 3)^2 + 1 \)[/tex]