Question 58·Medium·Advanced Math

The coefficient of $x^2$ is positive, so the graph opens upward. Its vertex, the point where it turns, is its lowest point. That’s where the minimum occurs.

Type $f(x)=2x^2-12x+7$. Click the curve, then click the gray point at its bottom. Desmos shows $(3,-11)$. The first coordinate is the input; the second coordinate is $f(x)$. The minimum is the vertex’s output, not its input. So the minimum value of $f(x)$ is $-11$. Choice B.

Factor $2$ out of the terms containing $x$, leaving the constant alone. This makes the expression ready for a perfect square:

Half of $-6$ is $-3$, and its square is $9$. Adding $9$ inside $2(\cdots)$ adds $18$ to the function, so subtract $18$ outside to keep the value unchanged:

The three terms inside the parentheses form $(x-3)^2$. Rewrite them as a square: $f(x)=2(x-3)^2+7-18$. Combine the constants: $f(x)=2(x-3)^2-11$. This vertex form makes the lowest possible output visible.

A square is never negative, so $2(x-3)^2\ge 0$ for every real $x$. At $x=3$, the square equals $0$. So the minimum value of $f(x)$ is $-11$. Choice B.