Question 198545
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Given: {{{system(5x^2=4x^2-4x)}}}


Continuing,
{{{5x^2-4x^2=4x=0}}}
{{{x^2+4x=0}}}


We complete the square to get the vertex;
{{{(x^2+4x+highlight(4))-highlight(4)=0}}}, take half of the middle constant, then squared = {{{(1/2)(4)=2^2=highlight(4)}}}


Then "add" and also "subtract" to the eqn, so it won't affect the operation:


{{{(x+2)^2-4=0}}}  via vertex form, {{{y=a(x-h)^2+k}}}, where{{{system(h=red(x),k=red(y))}}}

So the vertex, {{{red(x=-2)}}}, {{{red(y=-4)}}} ---> (-2,-4)



For the x-Intercept:
Continuing the Eqn,
{{{x^2+4x=0}}}
{{{x(x+4)=0}}}
{{{red(x=0)}}}; {{{red(x=-4)}}}


{{{drawing(300,300,-10,10,-10,10,grid(1),graph(300,300,-10,10,-10,10,x^2+4x),blue(circle(0,0,.25)),blue(circle(-4,0,.25)),red(circle(-2,-4,.25)))}}}


Thank you,
Jojo</font>