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Advanced Math / Nonlinear functions Difficulty: Hard

$f (x) = (x - 2) (x + 15)$

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

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Explanation

The correct answer is $- \frac{13}{2}$ . The value of $x$ for which $f (x)$ reaches its minimum can be found by rewriting the given equation in the form $f (x) = {(x - h)}^{2} + k$ , where $f (x)$ reaches its minimum, $k$ , when the value of $x$ is $h$ . The given equation, $f (x) = (x - 2) (x + 15)$ , can be rewritten as $f (x) = x^{2} + 13 x - 30$ . By completing the square, this equation can be rewritten as $f (x) = (x^{2} + 13 x + {(\frac{13}{2})}^{2}) - 30 - {(\frac{13}{2})}^{2}$ , which is equivalent to $f (x) = {(x + \frac{13}{2})}^{2} - \frac{289}{4}$ , or $f (x) = {(x - (- \frac{13}{2}))}^{2} - \frac{289}{4}$ . Therefore, $f (x)$ reaches its minimum when the value of $x$ is $- \frac{13}{2}$ . Note that -13/2 and -6.5 are examples of ways to enter a correct answer.

Alternate approach: The graph of $y = f (x)$ in the xy-plane is a parabola. The value of $x$ for the vertex of a parabola is the x-value of the midpoint between the two x-intercepts of the parabola. Since it's given that $f (x) = (x - 2) (x + 15)$ , it follows that the two x-intercepts of the graph of $y = f (x)$ in the xy-plane occur when $x equals 2$ and $x = - 15$ , or at the points $left parenthesis 2 comma 0 right parenthesis$ and $left parenthesis negative 15 comma 0 right parenthesis$ . The midpoint between two points, $left parenthesis x 1 comma y 1 right parenthesis$ and $left parenthesis x 2 comma y 2 right parenthesis$ , is $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ . Therefore, the midpoint between $left parenthesis 2 comma 0 right parenthesis$ and $left parenthesis negative 15 comma 0 right parenthesis$ is $(\frac{2 - 15}{2}, \frac{0 + 0}{2})$ , or $(- \frac{13}{2}, 0)$ . It follows that $f (x)$ reaches its minimum when the value of $x$ is $- \frac{13}{2}$ . Note that -13/2 and -6.5 are examples of ways to enter a correct answer.