Question 1196808
 
given:

{{{y = 3x^2+15x+30}}}   AND   {{{y = -x^2+4x-8}}}


first solving for the x intercepts using the quadratic formula:

{{{y = 3x^2+15x+30}}}

{{{x=(-15+-sqrt(15^2-4*3*30))/(2*3) }}}

{{{x=(-15+-sqrt(225-360))/6 }}}

{{{x=(-15+-sqrt(-135))/6 }}}

{{{x=(-15+-i*sqrt(135))/6 }}}

solutions:

{{{x = -5/2 + (i*sqrt(15))/2}}}

{{{x = -5/2 - (i*sqrt(15))/2}}}

->if the solutions are complex, the quadratic does not have x-intercepts

see it on a graph:

{{{ graph( 600, 600, -10, 10, -10, 40,3x^2+15x+30) }}}



{{{y = -x^2+4x-8}}}


{{{x=(-4+-sqrt(4^2-4*(-1)*(-8)))/(2*(-1)) }}}

{{{x=(-4+-sqrt(16-32))/(-2) }}}

{{{x=(-4+-sqrt(-16))/(-2) }}}

{{{x=(-4+-4i)/(-2) }}}

{{{x=(-2+-2i)/(-1) }}}

{{{x=-(-2+-2i) }}}

solutions:

{{{x=-(-2+2i) }}} =>{{{x=2-2i }}}

or
{{{x=-(-2-2i) }}} =>{{{x=2+2i }}}

->if the solutions are complex, the quadratic does not have x-intercepts


see it on a graph:

{{{ graph( 600, 600, -10, 10, -10, 10, -x^2+4x-8) }}}



using factoring method:

remove the {{{constant}}} from the right side, factor what's left to determine{{{ x}}} intercepts


{{{0= 3x^2+15x}}}

{{{0 = 3x(x + 5)}}}

=>if {{{3x=0}}}=>{{{x=0}}}
=>if {{{x+5=0}}}=>{{{x=-5}}}


 The average of those is the {{{x}}} coordinate of the vertex.

{{{(0+(-5))/2=-5/2=-2.5}}}


substitute that back into the original equation to solve for the {{{y}}} coordinate of the vertex


{{{y= 3*(-2.5)^2+15*(-2.5)}}}
{{{y= -18.75}}}


so we have the vertex at ({{{-2.5}}},{{{-18.75}}})    

{{{ graph( 600, 600, -10, 10, -20, 40,3x^2+15x) }}}



do same with other equation

{{{0 = -x^2+4x}}}

{{{0 = -x(x-4)}}}

solutions:

if {{{0 = -x}}} =>{{{x=0}}}

if {{{0 = x-4}}}=>{{{x=4}}}


average is {{{(0+4)/2=2}}}


{{{y= -2^2+4*2}}}

{{{y= -4+8}}}

{{{y= 4}}}


so we have the vertex at ({{{2}}},{{{4}}})  


{{{ graph( 600, 600, -10, 10, -10, 40,-x^2+4x) }}}