Question 51310
Notice the THREE parts to this solution:

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CASE I:
Discriminant is positive:  {{{x^2-4 =0}}}  
There are TWO real solutions.  The implication for the graph of {{{y=x^2 -4}}} is that the graph crosses the x axis at TWO points.
{{{graph(300,300,-6,6,-6,6,x^2-4)}}}


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CASE II:
Discriminant is zero:  {{{x^2-4x+4 =0}}}  
There is ONLY ONE real solution.  The implication for the graph of {{{y=x^2 -4x+4}}} is that the graph touches the x axis at only one point, but it does NOT cross the x axis.
{{{graph(300,300,-6,6,-6,6,x^2-4x+4)}}}


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CASE III:
Discriminant is negative:  {{{x^2+4 =0}}}  
There are NO real solutions.  The implication for the graph of {{{y=x^2 +4}}} is that the graph never touches nor crosses the x axis.
{{{graph(300,300,-6,6,-6,6,x^2+4)}}}


This is a VERY important concept, especially with graphing calculators!!  Congratulations on an excellent question!!


R^2 at SCC