Question 1181358

{{{F(x) = a(x-h)^2 + k }}}

given:
({{{7}}}, {{{4}}}) and ({{{4}}}, {{{3}}})
({{{8}}}, {{{24}}})


use :
({{{7}}}, {{{4}}}) 


{{{4 = a(7-h)^2 + k }}}...solve for {{{k}}}

{{{4 -a(7-h)^2= k }}}............eq.1


use :({{{4}}}, {{{3}}})

{{{3 = a(4-h)^2 + k }}}...solve for {{{k}}}

{{{3 - a(4-h)^2 = k }}}.........eq.2


use : ({{{8}}}, {{{24}}})

{{{24 = a(8-h)^2 + k }}}...solve for {{{k}}}
{{{24 - a(8-h)^2 =k }}}..........eq.3


from eq.1 and eq.2 we have

{{{4 -a(7-h)^2=3 - a(4-h)^2 }}}.....solve for {{{a}}}

{{{4 -3=a(7-h)^2 - a(4-h)^2 }}}

{{{1=a((7-h)^2 - (4-h)^2) }}}

{{{a=1/(33 - 6h) }}}...........eq.1)



from eq.1 and eq.3 we have

{{{4 -a(7-h)^2=24 - a(8-h)^2}}}.....solve for {{{a}}}

{{{4 -24=a(7-h)^2 - a(8-h)^2}}}
{{{ -20=a((7-h)^2 - (8-h)^2)}}}
{{{ -20=a(2h - 15)}}}
{{{ a=-20/(2h - 15)}}}............eq.2)



from eq.1) and eq.2) we have

{{{1/(33 - 6h) =-20/(2h - 15)}}}......solve for {{{h}}}

{{{1(2h - 15) =-20(33 - 6h)}}}

{{{2h - 15 =-660 +120h}}}

{{{660- 15 = 120h-2h}}}

{{{645 = 118h}}}

{{{highlight(h=645/118)}}}



go to

{{{a=1/(33 - 6h) }}}...........eq.1), substitute {{{h}}}

{{{a=1/(33 - 6(645/118)) }}} 

{{{a=1/(12/59) }}} 

{{{highlight(a=59/12) }}} 


go to

{{{3 - a(4-h)^2 = k }}}.........eq.2, substitute {{{a}}} and {{{h}}}

{{{3 - (59/12 )(4-645/118)^2 = k }}}
{{{k=3 - (59/12 )(29929/13924)  }}}
{{{k=3 - 29929/2832  }}}
{{{highlight(k=-21433/2832 ) }}}



so, your equation is:


{{{F(x) = (59/12)(x-645/118)^2 -21433/2832}}}