Question 672681
Vertex form of a quadratic equation: {{{y=a(x-h)^2+k}}}

given: vertex ({{{-1}}},{{{4}}}) that contains ({{{2}}},{{{7}}}) 

- Plug in  the vertex and {{{x}}} and {{{y}}} coordinates of the point given, then solve for "{{{a}}}"

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

{{{7=a(2+1)^2+4}}}

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

{{{3=9a}}}

{{{3/9=a}}}

{{{1/3=a}}}

Now substitute "a" and the vertex into the vertex form:

{{{y=(1/3)(x+1)^2+4}}}

{{{y=(1/3)(x^2+2x+1)+4}}}

{{{y=(1/3)x^2+(2/3)x+1/3+4}}}

{{{y=(1/3)x^2+(2/3)x+13/3}}}

{{{drawing(600,600,-10,10,-10,10,grid(0),circle(-1,4,0.2),locate(-1,4-.2,"V(-1,4)"),circle(2,7,0.2),graph(600,600,-10,10,-10,10,(1/3)(x+1)^2+4))}}}