Question

Describe how the equation y=(x-2)^2+5 transforms the graph of a basic parabola y=x^2

Answer 1

We need to move right twice and up five.

Explanation:

We could compare the vertex.

The vertex form a parabola is
`y=a(x-h)^2+k`. This is called vertex form because it tells you the vertex
`(h,k)`. It can also tell you if has been reflected or vertically squished/stretched all by looking at
`a`.

`y=(x-2)^2+5` has vertex (2,5).

`y=x^2=(x-0)^2+0` has vertex (0,0).

Let's describe how to get from the point (0,0) to the point (2,5).

We need to move right twice and up five.

So that is the transformation that will get you from
`y=x^2` to
`y=(x-2)^2+5`.

Answer 2

Answer: The graph of the basic parabola
`y=x^2` is shifted upward 5 units and shifted 2 units to the right.

Explanation:

There are some transformations for a function f(x).

Some of this transformations are the following:

If
`f(x)+k`, then the function is shifted upward "k" units

If
`f(x)-k`, then the function is shifted downward "k" units.

If
`f(x+k)`, then the function is shifted to the left "k" units.

If
`f(x-k)`, then the function is shifted to the right "k" units.

In this case we know that the equation
`y=(x-2)^2+5` is obtained by transformating
`y=x^2`.

Therefore, based on the transformations mentioned before, we can describe how the equation
`y=(x-2)^2+5` transforms the graph of the basic parabola
`y=x^2`:

The graph of the basic parabola
`y=x^2` is shifted upward 5 units and shifted 2 units to the right.