Question

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

Answer 1

We can write your function in the general form: `y=ax^2+bx+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=-b/(2a)=-(-2)/(2*1)=1` you can substitute this value into your function to find the `y` coordinate:
`y_v=(1)^2-(2*1)=-1`
So vertex `V` at `(1,-1)`.

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^2-2x=0` that solved goves you:
`x(x-2)=0`
and: `x_1=0` and `x_2=2`.
So x-intercepts at `(0,0)` and `(2,0)`.

You can now plot your parabola using mainly these points: