A quadratic function is given.
(a) f(x) = -x^2 + 4x + 2 = -(x^2 - 4x) + 2 = -[(x-2)^2 -4] + 2 = -(x-2)^2 + 6. It is the standard form (perfect square is picked out).
the vertex is where the perfect square is zero, i.e. at x=2. For this x y=6. (x, y) = (2, 6).
the x-intercepts are roots, f(x) = -(x-2)^2 + 6 = 0. So (x-2)^2 = 6, x-2 = +-sqrt(6), x = 2 +- sqrt(6). y of course 0: (x, y) = (2-sqrt(6), 0) (smaller x-value)(x, y) = (2+sqrt(6), 0) (larger x-value)
(sqrt is for square root)
the y-intercept is at x=0, f(0) = 2.(x, y) = (0, 2).
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