How do you graph #y=x^2-2x+3#?

1 Answer
Dec 29, 2014

You can look at the "special" points of your function. These are points that characterize the curve represented by your function.
In your case you have a quadratic in the general form given as:
#y=ax^2+bx+c#
which is represented, graphically, by a PARABOLA. The orientation of the parabola is given by the coefficient #a# of #x^2#; in this case you have #a=1>0# so this is an upward parabola, i.e. like a U.

Now the special points:
1) You find the VERTEX (the lowest point of your parabola) which has coordinates given by:
#x_v=-b/(2a) and y_v=-Delta/(4a)#;
(Where #Delta=b^2-4ac#);
2) Y-axis intecept: The coordinates of this pont are given as: #(c,0)#;
3) X-axis intercept(s): the coordinate of these intercepts (if they exist) are given by putting #y=0# and solving the second degree equation:
#ax^2+bx+c=0#

In your case you have:
enter image source here
With only these points we can already plot our parabola:
enter image source here