Question 409132
a focus at (9,15) and 

a vertex at (9,8)


Since the {{{x-coordinates}}} of the {{{vertex}}} and {{{focus}}} are the {{{same}}}, they are one of top of the other, so this is a {{{regular}}} vertical parabola, where the {{{x }}}part is squared. 

Since the vertex is {{{below}}} the focus, this is a {{{right-side}}}{{{ up}}} parabola and {{{p }}}is positive. 

Since the vertex and focus are 15 –8 = 7 units apart, then {{{p = 7}}}. 

And that's all I need for my equation, since they already gave me the vertex.



    (x–h)^2=4p(y –k)

    (x–9)^2 = 4*7(y –8)

    (x–9)^2 = 28(y –8)

    (x–9)^2/28= y–8
 
   (x–9)^2/28 +8= y 


 {{{y =(x^2-18x +81)/28 +8 }}}

{{{y =(1/28)x^2-0.64x +2.9 +8 }}}

{{{y = (1/28)x^2-0.64x +10.9 }}}



{{{ graph( 600, 600, -35, 25, -35, 25, (1/28)x^2-0.64x +10.9) }}}