Question 314411
Sketch the graphs of y=x^2-4x+3 and x-2y=-6 on the same set of axes. Find the coordinates of each intersection point.
<pre><b>
Solve the system:

{{{system(y=x^2-4x+3,x-2y=-6)}}}

Substitute {{{(x^2-4x+3)}}} for y in the second equation:

{{{x-2y=-6)}}}
{{{x-2(x^2-4x+3)=-6)}}}
{{{x-2x^2+8x-6=-6)}}}
{{{-2x^2+9x-6=-6)}}}
{{{-2x^2+9x=0)}}}
{{{x(-2x+9)=0}}}

Set each factor = 0

x=0  -2x+9=0
       -2x=-9
         x=4.5

Find the y-value that goes with x=0 by
substituting 0 for x in

{{{x-2y=-6)}}}
{{{0-2y=-6)}}}
{{{-2y=-6)}}}
{{{y=3)}}}

So one point of intersection is (x,y) = (0,3)

Find the y-value that goes with x=4.5 by
substituting 4.5 for x in

{{{x-2y=-6)}}}
{{{4.5-2y=-6)}}}
{{{-2y=-10.5)}}}
{{{y=5.25)}}}

So the other point of intersection is (x,y) = (4.5,5.25)        

{{{drawing(400,400,-2,7,-3,6,
locate(4.5,5.25,"(4.5,5.25)"), locate(.1,3,"(0,3)"),
graph(400,400,-2,7,-3,6,x^2-4x+3),
graph(400,400,-2,7,-3,6,(-6-x)/(-2)) )}}}

Edwin</pre>