f x = x - 2 x + 15 The function f is defined by the

The correct answer is $-\frac{13}{2}$. The value of $x$ for which $f\left(x\right)$ reaches its minimum can be found by rewriting the given equation in the form $f\left(x\right)={\left(x-h\right)}^2+k$, where $f\left(x\right)$ reaches its minimum, $k$, when the value of $x$ is $h$. The given equation, $f\left(x\right)=\left(x-2\right)\left(x+15\right)$, can be rewritten as $f\left(x\right)=x^2+13x-30$. By completing the square, this equation can be rewritten as $f\left(x\right)=\left(x^2+13x+{\left(\frac{13}{2}\right)}^2\right)-30-{\left(\frac{13}{2}\right)}^2$, which is equivalent to $f\left(x\right)={\left(x+\frac{13}{2}\right)}^2-\frac{289}{4}$, or $f\left(x\right)={\left(x-\left(-\frac{13}{2}\right)\right)}^2-\frac{289}{4}$. Therefore, $f\left(x\right)$ 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 $y=f\left(x\right)$ 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 $f\left(x\right)=\left(x-2\right)\left(x+15\right)$, it follows that the two x-intercepts of the graph of $y=f\left(x\right)$ in the xy-plane occur when $x=2$ and $x=-15$, or at the points $\left(2,0\right)$ and $\left(-15,0\right)$. The midpoint between two points, $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$, is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. Therefore, the midpoint between $\left(2,0\right)$ and $\left(-15,0\right)$ is $\left(\frac{2-15}{2},\frac{0+0}{2}\right)$, or $\left(-\frac{13}{2},0\right)$. It follows that $f\left(x\right)$ 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.