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

1 Answer
Gió
Sep 23, 2015

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:
enter image source here

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