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

Sep 23, 2015

We can write your function in the general form: $y = a {x}^{2} + b x + c$ where, in your case: $a = 1$, $b = - 2$ and $c = 0$.

#### Explanation:

Basically, to graph your function (that is a Quadratic, representing a Parabola) you need to observe:

1) the coefficient $a$ of ${x}^{2}$ (i.e. the number in front of it).
If it is $a > 0$ your parabola will be in the shape of an U (upward concavity) otherwise it will be the other way round. In your case is in the U shape;

2) the vertex: this is the highest/lowest point reached by your parabola and the parabola is plotted all around it. The coordinates of this special point are given as:
${x}_{v} = - \frac{b}{2 a} = - \frac{- 2}{2 \cdot 1} = 1$ you can substitute this value into your function to find the $y$ coordinate:
${y}_{v} = {\left(1\right)}^{2} - \left(2 \cdot 1\right) = - 1$
So vertex $V$ at $\left(1 , - 1\right)$.

3) y-intercept. You can set $x = 0$ in your function to find $y = 0$

4) x-intercept(s) (when they exist). You set $y = 0$ in your function to get:
${x}^{2} - 2 x = 0$ that solved goves you:
$x \left(x - 2\right) = 0$
and: ${x}_{1} = 0$ and ${x}_{2} = 2$.
So x-intercepts at $\left(0 , 0\right)$ and $\left(2 , 0\right)$.

You can now plot your parabola using mainly these points: 