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Find the largest value of $c$ such that $1$ is in the range of $f(x)=x^2-5x+c$.

 Feb 21, 2021
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Complete the square: \(f(x) = x^2 - 5x + c = (x - \frac{5}{2})^2 + (c - \frac{25}{4})\)

 

So, \(c - \frac{25}{4} = 1\) which  makes the minimum value of c equal to 29/4.

 Feb 21, 2021

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