Question 952421
first you are looking for the quadratic equation {{{y=ax^2+bx+c}}}

since you are given roots {{{ x[1]=-1}}},{{{x[2]=5}}}, you will use zero product theorem to find equation:

 {{{(x-x[1])(x-x[2])=0}}}
 {{{(x-(-1))(x-5)=0}}}
{{{(x+1)(x-5)=0}}}

and passes through the point ({{{1}}},{{{1}}}),so we need a constant {{{c}}} to multiply zero product using given point 

{{{y=c(x+1)(x-5) }}} ...........now if you sub in {{{x=1}}} and {{{y=1}}}
{{{1=c(1+1)(1-5) }}}
{{{1=c(2)(-4) }}}
{{{1=-8c}}}
 
giving {{{c=-(1/8)}}}

and, we have

{{{y=-(1/8)(x+1)(x-5)}}}
{{{y=-(1/8)(x^2-4 x-5)}}}
{{{y=-(1/8)x^2+x/2+5/8}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,1,.12),locate(1,1,p(1,1)),
circle(5,0,.12),locate(5,0.5,r(5,0)),
circle(-1,0,.12),locate(-1.5,0.5,r(-1,0)),
 graph( 600, 600, -10, 10, -10, 10, -(1/8)x^2+x/2+5/8)) }}}