Question 732109
given:  vertex (-1,2 ) and focus (-1,3).
find: the standard equation for the parabola and graph it

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 {{{3 -2 = 1}}} units apart, then {{{p = 1}}}.

that's all we need for equation:
    {{{(x -h)^2 = 4p(y - k)}}}

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

    {{{(x +1)^2 = 4(y-2)}}}....vertex form

the standard form of a parabola in the form of 

{{{y = a x^2 + b x + c}}}

so,  expand and solve  {{{(x +1)^2 = 4(y-2)}}} for {{{y}}}

{{{x^2+2x +1 = 4y-8}}}

{{{x^2+2x +1+8 = 4y}}}

{{{x^2+2x +9 = 4y}}}

{{{x^2/4+2x/4 +9 /4= y}}}

{{{y=(1/4)x^2+(1/2)x +2.25}}}


graph:


 {{{ drawing( 600, 600, -10, 10, -5, 10,circle(-1,2,0.1),circle(-1,3,0.1),locate(-1.3,2.3-.2,"V"),locate(-1.3,3.3-.2,"F"),graph( 600, 600, -10, 10, -5, 10,(1/4)x^2+(1/2)x +2.25)) }}}