Question 1127545
{{{y= x^2-3x+4}}}....eq.1

{{{x+4=4}}}....eq.2
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{{{x+4=4}}}....eq.2...solve  for {{{x}}}

{{{x=4-4}}}

{{{x=0}}}....for any point on the y-axis, {{{x=0}}} ;
furthermore, if any point for which the x-coordinate is equal to zero will be on the y-axis

substitute {{{x=0}}} in eq.1


{{{y= 0^2-3*0+4}}}....eq.1

{{{y= 4}}}

solution: intersection point

({{{0}}},{{{ 4}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
blue(line(0,-10,0,10)),circle(0,4,.12),locate(0.2,4,p(0,4)),
 graph( 600, 600, -10, 10, -10, 10,x^2-3x+4)) }}}