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Problem 470

AMC 12 late, AIME early

Algebra Difficulty 4.9 Find the answer

5. The quadratic function $f(x)=a+b x-x^{2}$ satisfies $f(1+x)=f(1-x)$ for any real number $x$ , and $f(x+m)$ is an increasing function on $(-\infty, 4]$ . Then the range of the real number $m$ is $\qquad$ .

Your answer

Official solution

5. $m \in(-\infty,-3]$

From $f(1+x)=f(1-x)$ , we know that $x=1$ is the axis of symmetry of $f(x)$ , thus $m+4 \leqslant 1, m \leqslant-3$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.