# How do you identify the important parts of y= x^2 - 4x to graph it?

Oct 5, 2015

Vertex $\left(2 , - 4\right)$

Axis of symmetry $x = 2$
since the co-efficient of ${x}^{2}$ is positive, the curve is concave upwards. It has a minimum.

#### Explanation:

Given -
$y = {x}^{2} - 4 x$
It is a quadratic function of the form $y = {x}^{2} + b x + c$
In the given function the constant term is absent.
So we shall have it as '0'.
$y = {x}^{2} - 4 x + 0$
The presence or absence of constant term is not going to affect our answer.

Find the vertex
$x = \frac{- b}{2 a} = \frac{- \left(- 4\right)}{2 \times 1} = \frac{4}{2} = 2$
$y = {2}^{2} - 4 \left(2\right) = 4 - 8 = - 4$

Vertex $\left(2 , - 4\right)$

Axis of symmetry $x = 2$
since the co-efficient of ${x}^{2}$ is positive, the curve is concave upwards. It has a minimum.