设$a \ne b$且为实数,若二次函数$f(x)=x^2+ax+b$满足$f(a)=f(b)$,求$f(2)$
【答案】 解:
$$
\begin{aligned}
& \because f(a)=f(b) \\
& \therefore a^2+a^2+b=b^2+a b+b \\
& \therefore 2 a^2=b^2+a b \\
& f(x)=x^2+a x+b \text { 对称轴 } x=-\frac{a}{2} \\
\end{aligned}
$$

等价于
$$
\left\{\begin{array}{l}
a \rightarrow-\frac{a}{2} \\
b \rightarrow-\frac{a}{2}
\end{array}\right.
$$

$$
\begin{aligned}
\text { 即 } b-\left(-\frac{a}{2}\right) & =-\frac{a}{2}-a \\
\Rightarrow b+\frac{a}{2} & =-\frac{b}{2} a \therefore b=-2 a
\end{aligned}
$$

$$
\begin{aligned}
2 a^2 & =4 a^2-2 a^2 \text { 不可用 } \\
f(2) & =4+2 a+b \\
& =4+2 a-2 a \\
& =4 \\
\therefore f(2) & =4
\end{aligned}
$$





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