Question 926572
a quadratic function that has zero's at {{{x[1]=-3}}} and {{{x[2]=5 }}} and y-intercept at ({{{0}}},{{{30}}})

{{{y=ax^2+bx+c}}}...({{{0}}},{{{30}}})=>{{{c=30}}}


{{{y=ax^2+bx+30}}} if {{{x[1]=-3}}}=>{{{y=0}}}

{{{0=a(-3)^2+b(-3)+30}}}

{{{0=9a-3b+30}}}...reduce, divide by {{{3}}}

{{{0=3a-b+10}}}....eq.1


{{{y=ax^2+bx+30}}} if {{{x[1]=5}}}=>{{{y=0}}}

{{{0=a(5)^2+b(5)+30}}}

{{{0=25a+5b+30}}}...reduce, divide by {{{5}}}

{{{0=5a+b+6}}}....eq.2


solve the system:

{{{0=3a-b+10}}}....eq.1
{{{0=5a+b+6}}}....eq.2
_______________________add both

{{{0=3a-cross(b)+10+5a+cross(b)+6}}}

{{{0=3a+10+5a+6}}}

{{{0=8a+16}}}

{{{-16=8a}}}

{{{highlight(-2=a)}}}


plug it in eq.1 or 2 and find {{{b}}}

{{{0=3(-2)-b+10}}}....eq.1

{{{b=-6+10}}}

{{{highlight(b=4)}}}

so, your equation is: {{{y=-2x^2+4x+30}}}


{{{drawing( 600, 600, -10, 10, -10, 40,locate(0,30,p(0,30)),circle(0,30,.2),locate(5,-0.5,p(5,0)),locate(-3,-0.5,p(-3,0)),circle(5,0,.127),circle(-3,0,.127), graph( 600, 600, -10, 10, -10, 40, -2x^2+4x+30)) }}}