Given the quadratic function y=- x ^2+6x-5, when t ≤ x ≤ t+3, the maximum value of the function is m and the minimum value is n, m-n=3, then t=__
42+67-69
y = -x^2 + 6x - 5 = (5 - x)(x - 1)
Zeros y = 0 for x = 5 and x = 1
maximum at x = 3: y(3) = 4
assume t <= 3 <= t + 3
m = 4, n = 1
(5 - t) (t - 1) = 1
t^2 - 6t + 6 = 0
t = 3 - sqrt (3)
assumption is valid
@CoffeeBeanKiller said in #2:
> 42+67-69
= 40
easy!!!!!!!!!!!
@tpr said in #3:
> y = -x^2 + 6x - 5 = (5 - x)(x - 1)
> Zeros y = 0 for x = 5 and x = 1
> maximum at x = 3: y(3) = 4
> assume t <= 3 <= t + 3
> m = 4, n = 1
> (5 - t) (t - 1) = 1
> t^2 - 6t + 6 = 0
> t = 3 - sqrt (3)
> assumption is valid
t1=3-√3,t2=√3
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