Question

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

Answer 1

Answer:

We need to move right twice and up five.

Step-by-step 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.

Step-by-step 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.