Question 1137841


{{{y=x^2+8x+30}}}......e1.1
{{{y=5-2x }}}....eq.2

from eq.1 and eq.2 we have

{{{x^2+8x+30=5-2x }}}........solve for {{{x}}}

{{{x^2+8x+30-5+2x =0}}}

{{{x^2+10x+25 =0}}}

{{{x^2+10x+5^2 =0}}}

{{{(x + 5)^2 = 0}}

=>double solution

{{{x=-5}}}


go to {{{y=5-2x }}}....eq.2, substitute {{{x}}}


{{{y=5-2(-5) }}} 

{{{y=5+10 }}} 

{{{y=15}}} 


intersection point is at: ({{{-5}}},{{{15}}})



{{{drawing( 600, 600, -10, 10, -10, 20,

circle(-5,15,.13),locate(-5,15,p(-5,15)),
 graph( 600, 600, -10, 10, -10, 20, x^2+8x+30, 5-2x)) }}}