Question 1159347
Quadratic: 

{{{y=x^2-6x+8}}}......factor

{{{y=x^2-2x-4x+8}}}

{{{y=(x^2-2x)-(4x-8)}}}

{{{y=x(x-2)-4(x-2)}}}

{{{y=(x-4)(x-2)}}}


1). what are the x-intercepts?

set {{{y=0}}}

{{{0=(x-4)(x-2)}}}

if {{{(x-4)=0}}}->{{{x=4}}}
if {{{(x-2)=0}}}->{{{x=2}}}

the x-intercepts are: {{{x=2}}} and {{{x=4}}}


2). what is the domain?

{{{R}}} (all real numbers) (it's parabola)


3). what is the range?

since it's parabola that opens up, it's vertex is a minimum and range is all real numbers above y-coordinate of the vertex

so, rewrite equation in vertex form

{{{y=x^2-6x+8}}}.......complete square

{{{y=(x^2-6x)+8}}}

{{{y=(x^2-6x+b^2)-b^2+8}}}......{{{b=6/2=3}}}

{{{y=(x^2-6x+3^2)-3^2+8}}}

{{{y=(x-3)^2-9+8}}}

{{{y=(x-3)^2-1}}}-> {{{h=3}}} and {{{k=-1}}}-> vertex is at ({{{3}}},{{{-1}}})

so, the range is:

{ {{{y}}} element {{{R}}} : {{{y>=-1}}} }


 {{{drawing( 600, 600, -10, 10, -10, 10,
circle(3,-1,.12),locate(3,-1,v(3,-1)),
 graph( 600, 600, -10, 10, -10, 10, x^2-6x+8)) }}}