Question 956147

you are given 3 points which are: ({{{0}}},{{{0}}}), ({{{3.5}}},{{{4}}}) and ({{{7}}},{{{0}}})

you need a quadratic equation {{{y = ax^2 + bx + c}}} of a parabola which contain given points

so, use point ({{{0}}},{{{0}}})=> means {{{x = 0}}} and {{{y = 0}}}

{{{y = ax^2 + bx + c}}}
{{{0 = a*0^2 + b*0 + c}}}
{{{0 = 0 + 0 + c}}}------eq.1

=>{{{c=0}}}

so, we now that our equation will be {{{y = ax^2 + bx }}} and we need to find {{{a}}} and {{{b}}} 

now use ({{{3.5}}},{{{4}}})
{{{ x = 3.5}}}, {{{y = 4}}}
{{{y = ax^2 + bx }}}
{{{4 = a*3.5^2 + b*3.5 }}}
{{{4 = 12.25a + 3.5b }}}------eq.2.......solve for {{{a}}}

{{{ 12.25a=4-3.5b }}}

{{{ a=4/12.25-3.5b/12.25 }}}

{{{ a=0.33-0.28b }}}.........eq a1

now, we can use ({{{7}}},{{{0}}})

{{{0 = a*7^2 + b*7 }}}

{{{0 = 49a + 7b }}}.....solve for {{{a}}}

{{{ 49a=-7b }}}

{{{ a=-7b/49 }}}

{{{ a=-b/7 }}}...........eq a2


use eq a1 and eq a2, make right sides equal

{{{ 0.33-0.28b =-b/7}}} ....solve for {{{b}}}.......both sides multiply by {{{7}}}

{{{ 2.31-1.96b =-b}}}

{{{ 2.31 =1.96b-b}}}

{{{ 2.31 =0.96b}}}

{{{b= 2.31 /0.96}}}

{{{b= 2.4}}}

now find {{{ a=-b/7 }}}...........eq a2

{{{ a=-2.4/7 }}}

{{{ a=-0.34 }}}

{{{ y=-0.34x^2+2.4x }}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3.5,4.2,.12),circle(7,0,.12),locate(3.5,4.2,p(3.5,4)),
locate(7,0.2,p(7,0)),
 graph( 600, 600, -10, 10, -10, 10,-0.34x^2+2.4x)) }}}