[tex]\text{The vertex form of a parabola}\\\\y=a(x-h)^2+k\\\\\text{We have the vertex}\ (-2,\ 5)\to h=-2,\ k=5.\\\\\text{Substitute}\\\\y=a(x-(-2))^2+5=a(x+2)^2+5\\\\\text{We have the point}\ (5,\ -4).\\\text{Substitute the coordinates of the point to the equation:}\\\\-4=a(5+2)^2+5\\\\-4=a(7)^2+5\\\\-4=49a+5\qquad|-5\\\\-9=49a\qquad|:49\\\\a=-\dfrac{9}{49}\\\\Answer:\ \boxed{y=-\dfrac{9}{49}(x+2)^2+5}[/tex]
