Question 919360
{{{y=x^2-5x+6}}} .......1

{{{x+y=2}}}.......2
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{{{x+y=2}}}.......2.....plug in {{{y}}} from 1

{{{x+x^2-5x+6=2}}} ...........solve for {{{x}}}

{{{x^2-4x+6-2=0}}}

{{{x^2-4x+4=0}}}

{{{x^2-2x-2x+4=0}}}

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

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

{{{(x-2)(x-2)=0}}} => double solution {{{x=2}}}

go to

{{{y=x^2-5x+6}}} .......plug in {{{2}}} for {{{x}}} and solve for {{{y}}}

{{{y=2^2-5*2+6}}}

{{{y=4-10+6}}}

{{{y=10-10}}}

{{{y=0}}}
 
so,these lines intercept at ({{{2}}},{{{0}}}) and a tangent line to the curve {{{y=x^2-5x+6}}} is at {{{x=2}}},the point where the curve intersects the x-axis


{{{ graph( 600, 600, -10,10, -10, 10, -x+2, x^2-5x+6) }}}