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Advanced Math Difficulty: Hard

$f (x) = 4 x^{2} - 50 x + 126$

The given equation defines the function $f$ . For what value of $x$ does $f (x)$ reach its minimum?

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Explanation

The correct answer is $StartFraction 25 Over 4 EndFraction$ . The given equation can be rewritten in the form $f (x) = a {(x - h)}^{2} + k$ , where $a$ , $h$ , and $k$ are constants. When $a greater than 0$ , $h$ is the value of $x$ for which $f (x)$ reaches its minimum. The given equation can be rewritten as $f (x) = 4 (x^{2} - \frac{50}{4} x) + 126$ , which is equivalent to $f (x) = 4 (x^{2} - \frac{50}{4} x + {(\frac{50}{8})}^{2} - {(\frac{50}{8})}^{2}) + 126$ . This equation can be rewritten as $f (x) = 4 ({(x - \frac{50}{8})}^{2} - {(\frac{50}{8})}^{2}) + 126$ , or $f (x) = 4 {(x - \frac{50}{8})}^{2} - 4 {(\frac{50}{8})}^{2} + 126$ , which is equivalent to $f (x) = 4 {(x - \frac{25}{4})}^{2} - \frac{121}{4}$ . Therefore, $h = \frac{25}{4}$ , so the value of $x$ for which $f (x)$ reaches its minimum is $StartFraction 25 Over 4 EndFraction$ . Note that 25/4 and 6.25 are examples of ways to enter a correct answer.