Question 757121

{{{y=x^2+2x-8}}}.......1 and 

{{{y= 2x+1}}}..........2
______________here you have system to solve.....subtract 2 from 1

{{{y-y=x^2+2x-8-(2x+1)}}}

{{{0=x^2+2x-8-2x-1}}}

or

{{{x^2+cross(2x)-8-cross(2x)-1=0}}}

{{{x^2-9=0}}}

{{{x^2=9}}}

{{{x=sqrt(9)}}}

{{{x=3}}} or {{{x=-3}}}

now find {{{y}}}

{{{y= 2x+1}}}..........2..plug in {{{x=3}}} 

{{{y= 2*3+1}}}

{{{y= 7}}}

and {{{x=-3}}}

{{{y= 2(-3)+1}}}

{{{y= -6+1}}}

{{{y=-5}}}

so, intersection points are: ({{{3}}},{{{7}}}) and ({{{-3}}},{{{-5}}})

see it on a graph:


{{{ graph( 600, 600, -10,10, -10, 10, 2x+1, x^2+2x-8) }}}