OG详解-OG9 数学1 Q24

Choice A is correct. It’s given that g(x) = f(x+5). Since f(x) = 4x²+64x+262, it  follows that f(x+5) = 4(x+5)² +64(x+5)+262. Expanding the quantity (x+5)² in this equation yields f(x+5) = 4(x+25)²+64(x+5)+262. Distributing the 4 and the 64 yields f(x+5) = 4x²+40x+100+64x+320+262. Combining like terms yields f(x+5) = 4x² +104x+682. Therefore, g(x) = 4x² +104x+682.   For a quadratic function defined by an equation of the form g(x) = a(x-h)² +k, where a, h, and k are constants and a is positive, g(x) reaches its minimum, k, when the value of x is h. The equation g(x) = 4x²+104x+682 can be rewritten in this form by completing the square. This equation is equivalent to g(x) = 4(x² +26x)+682, or g(x) = 4(x² +26x+169-169)+682. This equation can be rewritten as g(x) = 4((x+13)² -169)+682, or g(x) = 4(x+13)² -4(169)+682, which is equivalent to g(x) = 4(x+13)² +6. This equation is in the form g(x) = a(x-h)² +k, where a = 4, h =-13, and k = 6. Therefore, g(x) reaches its minimum when the value of x is -13.
Choice B is incorrect. This is the value of x for which f(x), rather than g(x), reaches its minimum. Choice C is incorrect and may result from conceptual or calculation errors. Choice D is incorrect. This is the value of x for which f(x-5), rather than f(x+5), reaches its minimum.