Question 1048624
<u>SOLVING</u>


{{{x^2+3x-(5x+1)+2=0}}}
{{{x^2+3x-5x-1+2=0}}}
{{{x^2-2x+1=0}}}
{{{(x-1)^2=0}}}
intersecting coordinate {{{x=1}}}.
Corresponding vertical coordinate, {{{y=5x+1=5*1+1=6}}}, for the single point (1,6).




<u>GRAPHING</u>


The line is straightforward.  


the parabola can be converted into standard form for easy finding of roots and vertex.  OR, factoring could help, if possible.


{{{y=x^2+3x+2}}}
{{{y=(x+1)(x+2)}}}
Roots being -1 and -2, symmetric around  {{{x=-3/2}}};
y for the vertex, {{{(-3/2+1)(-3/2+2)=(2/2-3/2)(4/2-3/2)=(-1/2)(1/2)=-1/4}}}.
-
Vertex is at  (-3/2, -1/4).




{{{graph(400,400,-3,7,-3,7,x^2+3x+2,5x+1)}}}


A closer look at the intersection:
{{{graph(400,400,-1,3,2,7,x^2+3x+2,5x+1)}}}