What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + t^2 - t + 21.
= 2t2 -10t -15 min occurs at t = - b/2a = - -10/4 = 2.5 use this value of 't' in the equation to calculate the minimum value of the quadratic
+14903 What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + t^2 - t + 21.
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\(f(x)=t^2 -9t - 36 + t^2 - t + 21\\ \frac{df(x)}{dx}=2t-9+2t-1=0\\ 4t=10\\ \color{blue}t=2.5 \)
The value of t = 2.5 produces the smallest value of the quadratic t^2 -9t - 36 + t^2 - t + 21.
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