《2025年学霸题中题九年级数学上册人教版》

1. (2023·台州中考)抛物线$y = ax^{2}-a(a≠0)与直线y = kx交于A(x_{1},y_{1})$,$B(x_{2},y_{2})$两点,若$x_{1}+x_{2}＜0$,则直线$y = ax + k$一定经过 (D)
A. 第一、二象限
B. 第二、三象限
C. 第三、四象限
D. 第一、四象限

答案： D

2. (2023·衡阳中考)已知$m＞n＞0$,若关于$x的方程x^{2}+2x - 3 - m = 0的解为x_{1}$,$x_{2}(x_{1}＜x_{2})$,关于$x的方程x^{2}+2x - 3 - n = 0的解为x_{3}$,$x_{4}(x_{3}＜x_{4})$,则下列结论正确的是 (B)
A. $x_{3}＜x_{1}＜x_{2}＜x_{4}$
B. $x_{1}＜x_{3}＜x_{4}＜x_{2}$
C. $x_{1}＜x_{2}＜x_{3}＜x_{4}$
D. $x_{3}＜x_{4}＜x_{1}＜x_{2}$

答案： B

3. (荆门中考)如图,函数$y= \begin{cases}x^{2}-2x + 3(x＜2),\\-\dfrac{3}{4}x+\dfrac{9}{2}(x\geqslant2)\end{cases}$的图象由抛物线的一部分和一条射线组成,且与直线$y = m$($m$为常数)相交于三个不同的点$A(x_{1},y_{1})$,$B(x_{2},y_{2})$,$C(x_{3},y_{3})(x_{1}＜x_{2}＜x_{3})$。设$t= \dfrac{x_{1}y_{1}+x_{2}y_{2}}{x_{3}y_{3}}$,则$t$的取值范围是$\frac{3}{5} ＜ t ＜ 1$。

答案： $\frac{3}{5} ＜ t ＜ 1$

4. (兰州中考)如图,抛物线$y= \dfrac{1}{2}x^{2}-7x+\dfrac{45}{2}与x轴交于点A$,$B$,把抛物线在$x轴及其下方的部分记作C_{1}$,将$C_{1}向左平移得到C_{2}$,$C_{2}与x轴交于点B$,$D$,若直线$y= \dfrac{1}{2}x + m与C_{1}$,$C_{2}共有3$个不同的交点,则$m$的取值范围是$-\frac{29}{8} ＜ m ＜ -\frac{5}{2}$。

答案： $-\frac{29}{8} ＜ m ＜ -\frac{5}{2}$

5. (湘西中考)已知二次函数$y = -x^{2}+4x + 5及一次函数y = -x + b$,将该二次函数在$x轴上方的图象沿x轴翻折到x$轴下方,图象的其余部分不变,得到一个新图象(如图所示),当直线$y = -x + b与新图象有4$个交点时,$b$的取值范围是______$-\frac{29}{4} ＜ b ＜ -1$。

答案： $-\frac{29}{4} ＜ b ＜ -1$

6. (2024·日照中考)已知二次函数$y = -x^{2}+(2a + 4)x - a^{2}-4a$($a$为常数)。
(1)求证：不论$a$为何值,该二次函数图象与$x$轴总有两个公共点。
(2)当$a + 1\leqslant x\leqslant2a + 5(a\geqslant - 1)$时,该二次函数的最大值与最小值之差为$9$,求此时函数的解析式。
(3)若二次函数图象对称轴为直线$x = 1$,该函数图象与$x轴交于A$,$B$两点(点$A在点B$左侧),与$y轴交于点C$。点$C关于对称轴的对称点为D$,点$M为CD$的中点,过点$M的直线l$(直线$l不过C$,$D$两点)与二次函数图象交于$E$,$F$两点,直线$CE与直线DF相交于点P$。
①求证：点$P$在一条定直线上；
②若$S_{\triangle COP}= \dfrac{3}{5}S_{\triangle ABP}$,请直接写出满足条件的直线$l$的解析式,不必说明理由。
(1) 令$y = 0$,则$-x^{2} + (2a + 4)x - a^{2} - 4a = 0$,$\because \Delta = (2a + 4)^{2} - 4(a^{2} + 4a) = 16 ＞ 0$,$\therefore$不论$a$为何值,方程总有两个不相等的实数根,$\therefore$二次函数图象与$x$轴总有两个公共点.
(2) 由二次函数的解析式得$y = -[x - (a + 2)]^{2} + 4$,$\therefore$函数图象对称轴为直线$x = a + 2$,最大值为$4$.$\because a + 2 - (a + 1) = 1$,$2a + 5 - (a + 2) = a + 3 \geq 2$,$\therefore |a + 2 - (a + 1)| ＜ |2a + 5 - (a + 2)|$,$\therefore$当$x = 2a + 5$时,$y$取得最小值,最小值为$-[2a + 5 - (a + 2)]^{2} + 4 = -(a + 3)^{2} + 4$,$\therefore 4 - [-(a + 3)^{2} + 4] = 9$,解得$a = 0$或$a = -6$(舍去),$\therefore$二次函数的解析式为$y = -x^{2} + 4x$.
(3) ①对称轴为直线$x = a + 2 = 1$,$\therefore a = -1$,$\therefore$二次函数解析式为$y = -x^{2} + 2x + 3$.令$y = 0$,则$-x^{2} + 2x + 3 = 0$,解得$x = -1$或$x = 3$,则$A(-1, 0)$,$B(3, 0)$,令$x = 0$,则$y = 3$,则$C(0, 3)$,$\therefore D(2, 3)$,$M(1, 3)$.设$E(m, -m^{2} + 2m + 3)$,$F(n, -n^{2} + 2n + 3)$,由题意知$m \neq n$,且均不为$0$,$2$.设直线$EF$的解析式为$y = k_{1}x + b_{1}$,$\therefore \begin{cases} -m^{2} + 2m + 3 = k_{1}m + b_{1}, \\ -n^{2} + 2n + 3 = k_{1}n + b_{1}, \end{cases}$解得$\begin{cases} k_{1} = 2 - m - n, \\ b_{1} = mn + 3, \end{cases}$ $\therefore$直线$EF$的解析式为$y = (2 - m - n)x + mn + 3$.又$\because$直线$EF$过点$M(1, 3)$,$\therefore 3 = 2 - m - n + mn + 3$,即$mn = m + n - 2$(记为①式).同理设直线$CE$的解析式为$y = k_{2}x + b_{2}$,把$E(m, -m^{2} + 2m + 3)$,$(0, 3)$代入得$\begin{cases} -m^{2} + 2m + 3 = k_{2}m + b_{2}, \\ 3 = b_{2}, \end{cases}$解得$\begin{cases} k_{2} = -m + 2, \\ b_{2} = 3, \end{cases}$ $\therefore$直线$CE$的解析式为$y = (-m + 2)x + 3$(记为②式).同理得直线$DF$的解析式为$y = -nx + 2n + 3$(记为③式).
由②③式联立得$\begin{cases} y = (-m + 2)x + 3, \\ y = -nx + 2n + 3, \end{cases}$解得$\begin{cases} x = \frac{2n}{n - m + 2}, \\ y = \frac{-2nm + 4n}{n - m + 2} + 3, \end{cases}$
$\therefore P\left(\frac{2n}{n - m + 2}, \frac{-2nm + 4n}{n - m + 2} + 3\right)$.若点$P$在一条定直线上,设点$P$所在直线解析式为$y = kx + b$,代入点$P$的坐标得$\frac{-2nm + 4n}{n - m + 2} + 3 = k \cdot \frac{2n}{n - m + 2} + b$,将①式代入化简得$-5m + 5n + 10 = -bm + (2k + b)n + 2b$,由对应系数相等得$k = 0$,$b = 5$,$\therefore$点$P$所在直线解析式为$y = 5$,即点$P$在一条定直线上.
②直线$l$的解析式为$y = \frac{3}{2}x + \frac{3}{2}$或$y = -\frac{5}{6}x + \frac{23}{6}$.

答案：
(1) 令$y = 0$,则$-x^{2} + (2a + 4)x - a^{2} - 4a = 0$,$\because \Delta = (2a + 4)^{2} - 4(a^{2} + 4a) = 16 ＞ 0$,$\therefore$不论$a$为何值,方程总有两个不相等的实数根,$\therefore$二次函数图象与$x$轴总有两个公共点.
(2) 由二次函数的解析式得$y = -[x - (a + 2)]^{2} + 4$,$\therefore$函数图象对称轴为直线$x = a + 2$,最大值为$4$.$\because a + 2 - (a + 1) = 1$,$2a + 5 - (a + 2) = a + 3 \geq 2$,$\therefore |a + 2 - (a + 1)| ＜ |2a + 5 - (a + 2)|$,$\therefore$当$x = 2a + 5$时,$y$取得最小值,最小值为$-[2a + 5 - (a + 2)]^{2} + 4 = -(a + 3)^{2} + 4$,$\therefore 4 - [-(a + 3)^{2} + 4] = 9$,解得$a = 0$或$a = -6$(舍去),$\therefore$二次函数的解析式为$y = -x^{2} + 4x$.
(3) ①对称轴为直线$x = a + 2 = 1$,$\therefore a = -1$,$\therefore$二次函数解析式为$y = -x^{2} + 2x + 3$.令$y = 0$,则$-x^{2} + 2x + 3 = 0$,解得$x = -1$或$x = 3$,则$A(-1, 0)$,$B(3, 0)$,令$x = 0$,则$y = 3$,则$C(0, 3)$,$\therefore D(2, 3)$,$M(1, 3)$.设$E(m, -m^{2} + 2m + 3)$,$F(n, -n^{2} + 2n + 3)$,由题意知$m \neq n$,且均不为$0$,$2$.设直线$EF$的解析式为$y = k_{1}x + b_{1}$,$\therefore \begin{cases} -m^{2} + 2m + 3 = k_{1}m + b_{1}, \\ -n^{2} + 2n + 3 = k_{1}n + b_{1}, \end{cases}$解得$\begin{cases} k_{1} = 2 - m - n, \\ b_{1} = mn + 3, \end{cases}$ $\therefore$直线$EF$的解析式为$y = (2 - m - n)x + mn + 3$.又$\because$直线$EF$过点$M(1, 3)$,$\therefore 3 = 2 - m - n + mn + 3$,即$mn = m + n - 2$(记为①式).同理设直线$CE$的解析式为$y = k_{2}x + b_{2}$,把$E(m, -m^{2} + 2m + 3)$,$(0, 3)$代入得$\begin{cases} -m^{2} + 2m + 3 = k_{2}m + b_{2}, \\ 3 = b_{2}, \end{cases}$解得$\begin{cases} k_{2} = -m + 2, \\ b_{2} = 3, \end{cases}$ $\therefore$直线$CE$的解析式为$y = (-m + 2)x + 3$(记为②式).同理得直线$DF$的解析式为$y = -nx + 2n + 3$(记为③式).
由②③式联立得$\begin{cases} y = (-m + 2)x + 3, \\ y = -nx + 2n + 3, \end{cases}$解得$\begin{cases} x = \frac{2n}{n - m + 2}, \\ y = \frac{-2nm + 4n}{n - m + 2} + 3, \end{cases}$
$\therefore P\left(\frac{2n}{n - m + 2}, \frac{-2nm + 4n}{n - m + 2} + 3\right)$.若点$P$在一条定直线上,设点$P$所在直线解析式为$y = kx + b$,代入点$P$的坐标得$\frac{-2nm + 4n}{n - m + 2} + 3 = k \cdot \frac{2n}{n - m + 2} + b$,将①式代入化简得$-5m + 5n + 10 = -bm + (2k + b)n + 2b$,由对应系数相等得$k = 0$,$b = 5$,$\therefore$点$P$所在直线解析式为$y = 5$,即点$P$在一条定直线上.
②直线$l$的解析式为$y = \frac{3}{2}x + \frac{3}{2}$或$y = -\frac{5}{6}x + \frac{23}{6}$.
解析:$\because A(-1, 0)$,$B(3, 0)$,$C(0, 3)$,$\therefore AB = | - 1 - 3 | = 4$,$OC = 3$.$\because S_{\triangle COP} = \frac{3}{5}S_{\triangle ABP}$,$\therefore \frac{1}{2}OC \cdot |x_{P}| = \frac{3}{5} \times \frac{1}{2}AB \cdot y_{P}$,$\therefore \frac{1}{2} \times 3 |x_{P}| = \frac{3}{5} \times \frac{1}{2} \times 4y_{P}$,$\therefore |x_{P}| = \frac{4}{5}y_{P}$,由①知$y_{P} = 5$,$\therefore |x_{P}| = 4$,$\therefore x_{P} = \pm 4$.当$x_{P} = 4$时,$\frac{2n}{n - m + 2} = 4$,整理得$n = 2m - 4$.又$\because mn = m + n - 2$,$\therefore m(2m - 4) = m + 2m - 4 - 2$,整理得$2m^{2} - 7m + 6 = 0$,解得$m_{1} = \frac{3}{2}$,$m_{2} = 2$(不符合题意,舍去),$\therefore n = 2m - 4 = 2 \times \frac{3}{2} - 4 = -1$,$\therefore k_{1} = 2 - m - n = \frac{3}{2}$,$b_{1} = mn + 3 = \frac{3}{2}$,$\therefore$直线$l$的解析式为$y = \frac{3}{2}x + \frac{3}{2}$;当$x_{P} = -4$时,$\frac{2n}{n - m + 2} = -4$,整理得$n = \frac{2m - 4}{3}$.又$\because mn = m + n - 2$,$\therefore m \cdot \frac{2m - 4}{3} = m + \frac{2m - 4}{3} - 2$,整理得$2m^{2} - 9m + 10 = 0$,解得$m_{1} = \frac{5}{2}$,$m_{2} = 2$(不符合题意,舍去),$\therefore n = \frac{2m - 4}{3} = \frac{1}{3}$,$\therefore k_{1} = 2 - m - n = -\frac{5}{6}$,$b_{1} = mn + 3 = \frac{23}{6}$,$\therefore$直线$l$的解析式为$y = -\frac{5}{6}x + \frac{23}{6}$.综上所述,当$S_{\triangle COP} = \frac{3}{5}S_{\triangle ABP}$时,直线$l$的解析式为$y = \frac{3}{2}x + \frac{3}{2}$或$y = -\frac{5}{6}x + \frac{23}{6}$.