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Quadratic function has the generic form as: y=ax^2+bx+c
Our purpose is to find a, b and c.
With information given as:
y-intercept is (0,-6) then we have: -6=a(0)^2+b(0)+c
from there we have c=-6 (we found c)
vertex is (-2,2) then we have 2=a(-2)^2+b(-2)-6 or 2a-b=4
with vertex at (-2,2) we aslo have the line of symmetry as x=-2
meaning as x=-4, y=-6 then -6=a(-4)^2+b(-4)-6 or 4a-b=0 or b=4a
substitute b=4a into the equation 2a-b=4 we have 2a-4a=4 or a=-2 (we found a)
substitute a=-2 into the equation b=4a we find b=-8 (we found b)
Then our quadratic function is: y=-2x^2-8x-6