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Advanced Math / Nonlinear functions Difficulty: Hard

fx= x - 10 x + 13

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

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Explanation

Choice D is correct. It's given that fx=x-10x+13, which can be rewritten as  fx=x2+3x-130. Since the coefficient of the x2-term is positive, the graph of y=fx in the xy-plane opens upward and reaches its minimum value at its vertex. The x-coordinate of the vertex is the value of x such that fx reaches its minimum. For an equation in the form fx=ax2+bx+c, where a , b , and c are constants, the x-coordinate of the vertex is -b2a. For the equation fx=x2+3x-130, a=1, b=3, and c=-130. It follows that the x-coordinate of the vertex is -321, or -32. Therefore, fx reaches its minimum when the value of x is -32.

Alternate approach: 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 fx=x-10x+13, it follows that the two x-intercepts of the graph of y=fx in the xy-plane occur when x=10 and x=-13, or at the points 10,0 and -13,0. The midpoint between two points, x1,y1 and x2,y2, is x1+x22,y1+y22. Therefore, the midpoint between 10,0 and -13,0 is 10+-132,0+02, or -32,0. It follows that fx reaches its minimum when the value of x is -32.

Choice A is incorrect. This is the y-coordinate of the y-intercept of the graph of y=fx in the xy-plane.

Choice B is incorrect. This is one of the x-coordinates of the x-intercepts of the graph of y=fx in the xy-plane.

Choice C is incorrect and may result from conceptual or calculation errors.