Question

What is the equation of the parabola that has a vertex at `(6, 2)` and passes through point `(3,20)`?

Answer 1

`y=2(x-6)^2+2`

Explanation:

Given:
`color(white)("XXX")`Vertex at `(color(red)6,color(blue)2)`, and
`color(white)("XXX")`Additional point at `(3,20)`

If we assume the desired parabola has a vertical axis,
then the vertex form of any such parabola is
`color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b` with vertex at `(color(red)a,color(blue)b)`

Therefore our desired parabola must have the vertex form
`color(white)("XXX")y=color(green)m(x-color(red)6)^2+color(blue)2`

Furthermore we know that the "additional point" `(x,y)=(color(magenta)3,color(teal)20)`

Therefore
`color(white)("XXX")color(teal)20=color(green)m(color(magenta)3-color(red)6)^2+color(blue)2`

`color(white)("XXX")rArr 18=9color(green)m`

`color(white)("XXX")rArr color(green)m=2`

Plugging this value back into our earier version of the desired parabola, we get
`color(white)("XXX")y=color(green)2(x-color(red)6)^2+color(blue)2`

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If the axis of symmetry is not vertical:
[1] if it is vertical a similar process can be used working with the general form `x=m(y-b)^2+a`
[2] if it is neither vertical nor horizontal, the process becomes more involved (ask as a separate question if this is the case; in general you will need to know the angle of the axis of symmetry in order to develop an answer).