Question

What are the coordinates of the vertex of the parabola whose equation is `y = 3(x - 2)^2 + 5`?

Answer 1

The answer is: `V(2,5)`.

There are two ways.

First:

we can remember the equation of the parabola, given the vertex `V(x_v,y_v)` and the amplitude `a`:

`y-y_v=a(x-x_v)^2`.

So:

`y-5=3(x-2)^2` has vertex: `V(2,5)`.

Second:

we can make the counts:

`y=3(x^2-4x+4)+5rArry=3x^2-12x+17`

and, remembering that `V(-b/(2a),-Delta/(4a))`,

`V(-(-12)/(2*3),-(12^2-4*3*17)/(4*3))rArrV(2,5)`.

Answer 2

Vertex is `(2, 5)`

Method

Use the form: `(x - h)^2 = 4a(y - k)`

This parabola has vertex at `(h, k)`
And its major axis is along the `y-"axis"`

In our case we have, `y = 3(x - 2)^2 + 5`

`=> 3(x - 2)^2 = y - 5`

`=> (x - 2)^2 = 1/3(y - 5)`

So, the vertex is `(2, 5)`

Worthy of note
When the equation is of the form: `(y - k)^2 = 4a(x - h)`

The vertex is at `(h, k)` and the parabola lies along the `x-"axis"`