# How do you identify the important parts of y=(x-2)^2 to graph it?

Oct 2, 2015

Vertex is $\left(2 , 0\right)$
Axis of symmetry $x = 2$

#### Explanation:

$y = {\left(x - 2\right)}^{2}$

It is a quadratic function in vertex form
$y = a \left(x - h\right) + k$
Where -
$\left(h , k\right)$ is vertex
$x = h$ is axis of symmetry.

In our case there is no $k$ term. We shall have it as $0$

$y = {\left(x - 2\right)}^{2} + 0$

Co-ordinates of the vertex

$x = - 1 \left(h\right) = - 1 \left(- 2\right) = 2$
$y = k = 0$

Vertex is $\left(2 , 0\right)$
Axis of symmetry $x = 2$

Since $a$ is positive, the curve is concave upwards.
It has a minimum.