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1. 已知点$A(m,y_{1})$，$B(m + 4,y_{2})$，$C(x_{0},y_{0})$在二次函数$y = ax^{2}+4ax + c(a\neq0)$的图像上，且$C$为抛物线的顶点. 若$y_{0}\geqslant y_{2}＞y_{1}$，则$m$的取值范围是
$m ＜ -4$
.
答案：
1.$m ＜ -4$ 点拨：由题意，得抛物线的对称轴为直线$x = - \frac{4a}{2a} = - 2$.
$\because C$为抛物线的顶点，且$y_{0} \geq y_{2} ＞ y_{1}$，$\therefore$抛物线开口向下.$\because y_{2} ＞ y_{1}$，$A(m,y_{1})$，$B(m + 4,y_{2})$，$\therefore AB$的中点位于对称轴的左侧，$\therefore\frac{m + m + 4}{2} ＜ - 2$，解得$m ＜ - 4$.
$\because C$为抛物线的顶点，且$y_{0} \geq y_{2} ＞ y_{1}$，$\therefore$抛物线开口向下.$\because y_{2} ＞ y_{1}$，$A(m,y_{1})$，$B(m + 4,y_{2})$，$\therefore AB$的中点位于对称轴的左侧，$\therefore\frac{m + m + 4}{2} ＜ - 2$，解得$m ＜ - 4$.
2. 已知二次函数$y = x^{2}-2nx + 3(n＞0)$，点$A(m - 2,a)$，$B(4,b)$，$C(m,a)$都在这个二次函数的图像上，且$a ＜ b ＜ 3$，则$m$的取值范围是
$3 ＜ m ＜ 4$或$m ＞ 6$
.
答案：
2.$3 ＜ m ＜ 4$或$m ＞ 6$ 点拨：$\because y = x^{2} - 2nx + 3 = (x - n)^{2} - n^{2} + 3$，$\therefore$二次函数$y = x^{2} - 2nx + 3$的对称轴为直线$x = n$.$\because$点$A(m - 2,a)$，$C(m,a)$在二次函数$y = x^{2} - 2nx + 3$的图像上，$\therefore\frac{m - 2 + m}{2} = m - 1 = n$.$\because n ＞ 0$，$\therefore m - 1 ＞ 0$，
解得$m ＞ 1$.
$\because m - 2 ＜ m$，$\therefore$点$A$在对称轴左侧，点$C$在对称轴右侧，在$y = x^{2} - 2nx + 3$中，令$x = 0$，得$y = 3$，
$\therefore$抛物线$y = x^{2} - 2nx + 3$与$y$轴的交点坐标为$(0,3)$，
$\therefore(0,3)$关于对称轴直线$x = m - 1$的对称点为$(2m - 2,3)$.
$\because b ＜ 3$，$\therefore 4 ＜ 2m - 2$，解得$m ＞ 3$.
①当点$A(m - 2,a)$，$B(4,b)$都在对称轴左侧时.
$\because y$随$x$的增大而减小，且$a ＜ b$，
$\therefore 4 ＜ m - 2$，解得$m ＞ 6$，此时$m$满足的条件为$m ＞ 6$
②当点$A(m - 2,a)$在对称轴左侧，点$B(4,b)$在对称轴右侧时.$\because a ＜ b$，
$\therefore$点$B(4,b)$到对称轴直线$x = m - 1$的距离大于点$A(m - 2,a)$到对称轴直线$x = m - 1$的距离，$\therefore 4 - (m - 1) ＞ m - 1 - (m - 2)$，解得$m ＜ 4$，此时$m$满足的条件是$3 ＜ m ＜ 4$.
综上所述，$m$的取值范围是$3 ＜ m ＜ 4$或$m ＞ 6$.
解得$m ＞ 1$.
$\because m - 2 ＜ m$，$\therefore$点$A$在对称轴左侧，点$C$在对称轴右侧，在$y = x^{2} - 2nx + 3$中，令$x = 0$，得$y = 3$，
$\therefore$抛物线$y = x^{2} - 2nx + 3$与$y$轴的交点坐标为$(0,3)$，
$\therefore(0,3)$关于对称轴直线$x = m - 1$的对称点为$(2m - 2,3)$.
$\because b ＜ 3$，$\therefore 4 ＜ 2m - 2$，解得$m ＞ 3$.
①当点$A(m - 2,a)$，$B(4,b)$都在对称轴左侧时.
$\because y$随$x$的增大而减小，且$a ＜ b$，
$\therefore 4 ＜ m - 2$，解得$m ＞ 6$，此时$m$满足的条件为$m ＞ 6$
②当点$A(m - 2,a)$在对称轴左侧，点$B(4,b)$在对称轴右侧时.$\because a ＜ b$，
$\therefore$点$B(4,b)$到对称轴直线$x = m - 1$的距离大于点$A(m - 2,a)$到对称轴直线$x = m - 1$的距离，$\therefore 4 - (m - 1) ＞ m - 1 - (m - 2)$，解得$m ＜ 4$，此时$m$满足的条件是$3 ＜ m ＜ 4$.
综上所述，$m$的取值范围是$3 ＜ m ＜ 4$或$m ＞ 6$.
3. 在平面直角坐标系$xOy$中，$M(x_{1},y_{1})$，$N(x_{2},y_{2})$是抛物线$y = a(x - t)^{2}+k(a＞0)$上任意两点.
(1)若对于$x_{1}=0$，$t = 1$，有$y_{1}=y_{2}$，求$x_{2}$的值；
(2)若对于$0 ＜ x_{1}＜1$，$t + 1 ＜ x_{2}＜t + 2$，都不存在$y_{1}=y_{2}$，求$t$的取值范围.
(1)若对于$x_{1}=0$，$t = 1$，有$y_{1}=y_{2}$，求$x_{2}$的值；
(2)若对于$0 ＜ x_{1}＜1$，$t + 1 ＜ x_{2}＜t + 2$，都不存在$y_{1}=y_{2}$，求$t$的取值范围.
答案：
3.解：
(1)$\because t = 1$，$\therefore$抛物线$y = a(x - t)^{2} + k(a ＞ 0)$的对称轴为直线$x = t = 1$.
$\because y_{1} = y_{2}$，$\therefore M(x_{1},y_{1})$，$N(x_{2},y_{2})$关于对称轴对称，
$\therefore\frac{x_{1} + x_{2}}{2} = 1$.
$\because x_{1} = 0$，$\therefore x_{2} = 2$.
(2)设$N(x_{2},y_{2})$关于直线$x = t$的对称点为$N^{\prime}(x_{2}^{\prime},y_{2})$.
$\therefore\frac{x_{2} + x_{2}^{\prime}}{2} = t$，$\therefore x_{2}^{\prime} = 2t - x_{2}$.
$\because t + 1 ＜ x_{2} ＜ t + 2$，$\therefore t - 2 ＜ x_{2}^{\prime} = 2t - x_{2} ＜ t - 1$.
$\because$不存在$y_{1} = y_{2}$，且$0 ＜ x_{1} ＜ 1$，$\therefore t - 2 \geq 1$或$t + 2 \leq 0$
或$\begin{cases}1 \leq t + 1, \\0 \geq t - 1. \end{cases}$
解得$t \geq 3$或$t \leq - 2$或$0 \leq t \leq 1$.
(1)$\because t = 1$，$\therefore$抛物线$y = a(x - t)^{2} + k(a ＞ 0)$的对称轴为直线$x = t = 1$.
$\because y_{1} = y_{2}$，$\therefore M(x_{1},y_{1})$，$N(x_{2},y_{2})$关于对称轴对称，
$\therefore\frac{x_{1} + x_{2}}{2} = 1$.
$\because x_{1} = 0$，$\therefore x_{2} = 2$.
(2)设$N(x_{2},y_{2})$关于直线$x = t$的对称点为$N^{\prime}(x_{2}^{\prime},y_{2})$.
$\therefore\frac{x_{2} + x_{2}^{\prime}}{2} = t$，$\therefore x_{2}^{\prime} = 2t - x_{2}$.
$\because t + 1 ＜ x_{2} ＜ t + 2$，$\therefore t - 2 ＜ x_{2}^{\prime} = 2t - x_{2} ＜ t - 1$.
$\because$不存在$y_{1} = y_{2}$，且$0 ＜ x_{1} ＜ 1$，$\therefore t - 2 \geq 1$或$t + 2 \leq 0$
或$\begin{cases}1 \leq t + 1, \\0 \geq t - 1. \end{cases}$
解得$t \geq 3$或$t \leq - 2$或$0 \leq t \leq 1$.
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