Question 58544
for the function (standard form){{{y=x^2-6x+8}}} put the function in the form (vertex form) {{{y=a(x-h)^2 +k}}}
{{{y=x^2-6x+8}}}  Group the x's
{{{y=(x^2-6x)+8}}}  Add (-6/2)^2=(-3)^2=9 to the inside of the parenthesis and take it away from the outside, so you're adding 9-9=0 and not breaking any rules.
{{{y=(x^2-6x+9)-9+8}}}  Now the parenthesis is a perfect square.
{{{y=(x-3)^2-1}}} 
a=1  since it's positive and 1 the parabola opens up and is standard width.
The vertex is (h,k)=(3,-1)  
So you have a general idea of where it is and what it looks like, but if you want to be extra careful you can plot the x and y intercepts using the standard form.
{{{y=0^2-6(0)+8}}} 
{{{y=8}}}  (0,8) is the y-intercept.
{{{0=x^2-6x+8}}} Factor
{{{0=(x-2)(x-4)}}}
x-2=0  and  x-4=0
x=2  and x=4  (2,0) and (4,0) are the x-intercepts.
This is what the graph looks like:
{{{graph(300,200,-10,10,-10,10,x^2-6x+8)}}}
Happy Calculating!!!