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Advanced Math / Nonlinear functions Difficulty: Hard

What is the minimum value of the function f defined by $f of x equals, open parenthesis, x minus 2, close parenthesis, squared, minus 4$ ?

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Explanation

Choice A is correct. The given quadratic function f is in vertex form, $f of x equals, open parenthesis, x minus h, close parenthesis, squared, plus k$ , where $the point with coordinates h comma k$ is the vertex of the graph of $y equals f of x$ in the xy-plane. Therefore, the vertex of the graph of $y equals f of x$ is $the point with coordinates 2 comma negative 4$ . In addition, the y-coordinate of the vertex represents either the minimum or maximum value of a quadratic function, depending on whether the graph of the function opens upward or downward. Since the leading coefficient of f (the coefficient of the term $open parenthesis, x minus 2, close parenthesis, squared$ ) is 1, which is positive, the graph of $y equals f of x$ opens upward. It follows that at $x equals 2$ , the minimum value of the function f is $negative 4$ .

Choice B is incorrect and may result from making a sign error and from using the x-coordinate of the vertex. Choice C is incorrect and may result from using the x-coordinate of the vertex. Choice D is incorrect and may result from making a sign error.