Question 120077
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} is the quadratic formula.  The portion of the formula under the radical ({{{b^2-4ac}}}) is called the <i>discriminant</i> because it discriminates as to the type of roots for a given set of coefficients a, b, and c.


if {{{b^2-4ac>0}}}, then the two roots of {{{ax^2+bx+c=0}}} are real and unequal.


if {{{b^2-4ac=0}}}, then the two roots of {{{ax^2+bx+c=0}}} are real and equal.


if {{{b^2-4ac<0}}}, then the two roots of {{{ax^2+bx+c=0}}} are a complex conjugate pair.


In the problem given, {{{a=1}}}, {{{b=4}}}, and {{{c=4}}}, so {{{b^2-4ac=4^2-4(1)(4)=16-16=0}}}.  This means that the two roots of {{{x^2+4x+4=0}}} are real and equal, therefore the graph of {{{f(x)=x^2+4x+4}}} intersects the x axis at one point -- because both of the roots of the quadratic equation exist at the same point.


The following illustrates the graphs of {{{f(x)=x^2+4x+3}}} (red graph, discriminant positive), {{{f(x)=x^2+4x+4}}} (green graph, discriminant zero), and {{{f(x)=x^2+4x+5}}} (blue graph, discriminant negative)


{{{graph(600,600,-5,5,-5,5,x^2+4x+3,x^2+4x+4,x^2+4x+5)}}}


Hope this helps,
John