What is the equation of a parabola that has a vertical axis, passes through the point [tex]\((-1,3)\)[/tex], and has its vertex at [tex]\((3,2)\)[/tex]?

A. [tex]\(x=-\frac{y^2}{16}+\frac{6 y}{16}-\frac{41}{16}\)[/tex]
B. [tex]\(y=-\frac{x^2}{16}+\frac{6 x}{16}-\frac{41}{16}\)[/tex]
C. [tex]\(y=\frac{x^2}{16}-\frac{6 x}{16}+\frac{41}{16}\)[/tex]
D. [tex]\(x=\frac{y^2}{16}-\frac{6 y}{16}+\frac{41}{16}\)[/tex]



Answer :

To determine the correct equation of the parabola that has a vertical axis, passes through the point [tex]\((-1,3)\)[/tex], and has its vertex at [tex]\((3,2)\)[/tex], we will use the given transformations and properties of parabolas.

1. Vertex Form of a Parabola:
The general form of a parabolic equation with a vertical axis and vertex at [tex]\((h, k)\)[/tex] is:
[tex]\[ (x - h)^2 = 4a(y - k). \][/tex]
This represents the parabola opening either upwards or downwards.

2. Given Information:
- Vertex: [tex]\((3, 2)\)[/tex]
- Point: [tex]\((-1, 3)\)[/tex]

3. Written in Standard Form:
We rearrange the terms to:
[tex]\[ (x - 3)^2 = 4a(y - 2) \][/tex]

4. Substitute the Given Point [tex]\((-1, 3)\)[/tex]:
To find [tex]\( a \)[/tex], substitute the point [tex]\((-1, 3)\)[/tex]:
[tex]\[ (-1 - 3)^2 = 4a(3 - 2) \][/tex]
Simplifying, we get:
[tex]\[ 16 = 4a \cdot 1 \][/tex]
Thus, [tex]\( a = 4 \)[/tex].

5. Equation of the Parabola:
Thus, the specific equation is:
[tex]\[ (x - 3)^2 = 16(y - 2) \quad \quad [1] \][/tex]
Rearranging to make [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y - 2 = \frac{(x - 3)^2}{16} \implies y = \frac{(x - 3)^2}{16} + 2 \][/tex]

6. Expanding the Final Formula:
Expanding [tex]\((x-3)^2\)[/tex]:
[tex]\[ y = \frac{x^2 - 6x + 9}{16} + 2 \][/tex]
[tex]\[ y = \frac{x^2}{16} - \frac{6x}{16} + \frac{9}{16} + 2 \][/tex]
[tex]\[ y = \frac{x^2}{16} - \frac{6x}{16} + \frac{9}{16} + \frac{32}{16} \][/tex]
[tex]\[ y = \frac{x^2}{16} - \frac{6x}{16} + \frac{41}{16} \][/tex]
Thus, the final equation of the parabola in standard form is:
[tex]\[ y = \frac{x^2}{16} - \frac{6x}{16} + \frac{41}{16} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{y = \frac{x^2}{16} - \frac{6 x}{16} + \frac{41}{16}} \][/tex]