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(x)= -x^2-3x+6 find the vertex, axis of symmetry, max or min value, range, interval increasing and decreasing, solve the equation f(x)=0
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f(x) = - x^2 - 3x + 6 . a) y-intercept --> x = 0 -> y = 6 x- intercepts -> y = 0 -> Use formula to solve y = -x^2 - 3x + 6 = 0 D = 9 + 24 = 32 = 16*2 --> VD = 4V2 and VD = -4V2 2 real roots: x1 = -3/2 + 4V2/2 = (-3 + 4V2)/2 = 1.32, and x2 = (-3 - 4V2)/2 = -4.32 b) The axis is at x = -b/2a = 3/-2 = -3/2 Since a is negative the parabola is downward. There is a max with vertex at x = -3/2 y of vertex: y (ver.) = -(9/4) + 9/2 + 6 = 33/4 c) Function y increases from -infinity to x = -3/2 Function y decreases from x =-3/2 to -infinity
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