Question 1189918

The graph of a certain quadratic 
{{{y = ax^2 + bx + c }}}

in vertex form"
 
{{{y = (ax^2 + bx) + c }}}

{{{y = a(x^2 + (b/a)x+(b/(2a))^2) -a(b/(2a))^2+ c }}}

{{{y = a(x^2 + (b/a)x+(b/(2a))^2) -b^2/4a+ c }}}

{{{y= a(x - b/(2a))^2 -b^2/4a+ c }}}


 a parabola with vertex ({{{-4}}},{{{0}}}) =>{{{h=-4}}}->{{{h=b/(2a)}}}=>{{{b/(2a)=-4}}}->{{{b=-8a}}}

{{{ k=0}}}->{{{0=-b^2/(4a)+ c }}}

{{{c=b^2/4a }}}->{{{c=(-8a)^2/(4a)}}}->{{{c=64a^2/(4a)}}}->{{{c=16a}}}


{{{y= a(x +4)^2 -(-8a)^2/(4a)+ 16a}}}

{{{y= a(x +4)^2 -16a+ 16a}}}

{{{y= a(x +4)^2  }}}


which passes through the point ({{{1}}},{{{-75}}}) 

{{{-75= a(1 +4)^2}}}
 
{{{-75= 25a}}}

{{{a=-3}}}


then your equation is:

{{{y=-3(x +4)^2}}}

{{{y=-3x^2 - 24x - 48}}}



{{{ drawing( 600, 600, -10, 10, -80, 10, 
circle(1,-75,.13),locate(1,-75,p(1,-75)),
circle(-4,0,.13),locate(-4,0,V(-4,0)),
graph( 600, 600, -10, 10, -80, 10, -3x^2 - 24x - 48)) }}}