let's find the equation in its standard form using the given information
since the vertex is (4,-10)
it must be
y = a(x-4)^2 -10
but (2,-6) lies on it
-6 = a(-2)^2 - 10
4 = 4a
a = 1
so y = (x-4)^2 - 10
y = x^2 - 8x + 16 - 10
y = x^2 - 8x + 6
so a=1, b=-8 and c=6
Hello. Please, help me solve this:
Parabola y = ax^2 + bx + c passes through the point (2;-6). Find the values of a, b and c, if vertex of parabola is (4;-10).
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1 answer