Question 198500
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Given:{{{system(Y=x^2-2x-1(red(EQN1)),Y=-3x+5(red(EQN2)))}}}



In {{{red(EQN1)}}}, it's a parabola and follows the form {{{Ax^2+Bx+C=0}}}


For the Vertex, {{{x=-b/2a=-(-2)/(2*1)=2/2=red(1)}}}, 
Substitute to the Eqn to  Y:
{{{Y=1^2-2(1)-1=1-2-1=red(-2)}}}


The Vertex ----> <font color=red>(1,-2)</font>


We solve the eqn to find the x-Intercepts:
{{{x^2-2x-1=0}}}


For the Discriminant: {{{b^2-4ac=-2^2-4(1)(-1)=4+4=8}}}


Then,
{{{x=(-(-2)+-sqrt(8))/(2*1)=(2+-2.83)/2}}}
{{{x=(2+2.83)/2=4.83/2=red(2.415)}}}
{{{x=(2-2.83)/2=-.83/2=red(-0.415)}}}


We see the graph,
{{{drawing(300,300,-5,5,-5,5,grid(1),graph(300,300,-5,5,-5,5),blue(circle(2.415,0,.12)),blue(circle(-.415,0,.12)),red(circle(1,-2,.12)))}}}--->{{{drawing(300,300,-5,5,-5,5,grid(1),graph(300,300,-5,5,-5,5,x^2-2x-1),blue(circle(2.415,0,.12)),blue(circle(-.415,0,.12)),red(circle(1,-2,.12)))}}}



In {{{red(EQN2)}}}, is a Line with  the form {{{Ax+By+C=0}}}.


{{{Y=-3x+5}}}
Let Fy=0
{{{0=-3x+5}}} ---> {{{3x=5}}} ----> {{{cross(3)x/cross(3)=5/3}}}
{{{red(x=5/3)}}}, x-Intercept (5/3,0) = (1.67,0)


Let Fx=0
{{{y=-3(0)+5}}}
{{{red(y=5)}}}, y-Intercept (0,5)


{{{drawing(300,300,-5,5,-5,10,grid(1),graph(300,300,-5,5,-5,10),blue(circle(5/3,0,.12)),blue(circle(0,5,.12)))}}}--->{{{drawing(300,300,-5,5,-5,10,grid(1),graph(300,300,-5,5,-5,10,-3x+5),blue(circle(5/3,0,.12)),blue(circle(0,5,.12)))}}}


Thank you,
Jojo</font>