Question 189678
{{{f(x)=a(x-h)^2+k}}} Start with the general vertex form.



{{{f(x)=a(x-0)^2+(-5)}}} Plug in {{{h=0}}} and {{{k=-5}}}



{{{f(x)=ax^2-5}}} Simplify


-------------------



{{{2=a(-5)^2-5}}} Plug in {{{x=-5}}} and {{{f(x)=2}}}



Note: since the graph passes through the point (-5,2), this means that if {{{x=-5}}}, then {{{f(x)=2}}} (f(x) is really "y")



{{{2=a(25)-5}}} Square -5 to get 25



{{{2=25a-5}}} Rearrange the terms



{{{2+5=25a}}} Add 5 to both sides.



{{{7=25a}}} Combine like terms.



{{{7/25=a}}} Divide both sides by 25 to isolate "a".



{{{a=7/25}}} Rearrange the equation



---------------


{{{f(x)=ax^2-5}}} Go back to the previous function



{{{f(x)=(7/25)x^2-5}}} Plug in {{{a=7/25}}} 





So the function with a vertex of (0,-5) and passes through (-5,2) is {{{f(x)=(7/25)x^2-5}}}