《2025年多维导学案九年级数学全一册北师大版》

1. 抛物线$y = 2x^2$向上平移1个单位长度得到的抛物线的解析式为$y = 2x^2 + 1$.

答案： $y = 2x^2 + 1$

2. 猜想:把抛物线$y = 2x^2$向左平移1个单位长度得到的抛物线的解析式为$y = 2(x + 1)^2$.

答案： $y = 2(x + 1)^2$

3. 在下面的平面直角坐标系中画出$y = 2(x + 1)^2$和$y = 2(x - 1)^2$的图象,并填表.

观察图象填空:
(1) 抛物线$y = 2x^2$向____平移____个单位长度可得到抛物线$y = 2(x + 1)^2$.
(2) 抛物线$y = 2x^2$向____平移____个单位长度可得到抛物线$y = 2(x - 1)^2$.

答案：
| $x$ | $\cdots$ | $-3$ | $-2$ | $-1$ | $0$ | $1$ | $2$ | $3$ | $\cdots$ |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| $y = 2x^2$ | $\cdots$ | $18$ | $8$ | $2$ | $0$ | $2$ | $8$ | $18$ | $\cdots$ |
| $y = 2(x + 1)^2$ | $\cdots$ | $8$ | $2$ | $0$ | $2$ | $8$ | $18$ | $32$ | $\cdots$ |
| $y = 2(x - 1)^2$ | $\cdots$ | $32$ | $18$ | $8$ | $2$ | $0$ | $2$ | $8$ | $\cdots$ |

(1)左 1
(2)右 1
| | 开口方向 | 对称轴 | 顶点坐标 |
| --- | --- | --- | --- |
| $y = 2(x + 1)^2$ | 向上 | 直线$x = -1$ | $(-1, 0)$ |
| $y = 2(x - 1)^2$ | 向上 | 直线$x = 1$ | $(1, 0)$ |

| | 开口方向 | 对称轴 | 顶点坐标 | 增减性和最值 |
| --- | --- | --- | --- | --- |
| $y = ax^2$ | $a ＞ 0$,开口向上 | $y$轴 | $(0,0)$ | 由前三个要素画出抛物线的大致图象,根据图象即可确定增减性和最值. |
| $y = a(x - h)^2$ | $a ＜ 0$,开口向下 | 直线$x = h$ | $(h, 0)$ | |

答案： 上下 直线$x = h$ $(h, 0)$

【例1】(1) 抛物线$y = x^2$向____平移____个单位长度,可得到抛物线$y = (x + 3)^2$;
(2) 抛物线$y = x^2$向____平移____个单位长度,可得到抛物线$y = (x - 3)^2$.

答案：

(1)左 3

(2)右 3

【变式1】(1) 抛物线$y = -2x^2$向左平移4个单位长度可得到抛物线的解析式为____.
(2) 抛物线$y = -\frac{1}{3}x^2$向右平移$\sqrt{3}$个单位长度可得到抛物线的解析式为____.

答案：

(1)$y = -2(x + 4)^2$

(2)$y = -\frac{1}{3}(x - \sqrt{3})^2$

【例2】已知抛物线$y = 5(x + 2)^2$;
(1) 开口向____;
(2) 顶点坐标是____;
(3) 对称轴是____;
(4) 当$x =$____时,$y$的最小值是____;
(5) 当$x$____时,$y$随$x$的增大而增大.

答案：

(1)上

(2)$(-2, 0)$

(3)直线$x = -2$

(4)$-2$ $0$

(5)$＞ -2$

【变式2】(1) 抛物线$y = -\sqrt{3}(x - 5)^2$的对称轴是____.
(2) 已知抛物线$y = -7(x + 10)^2$,下列说法错误的是 ( )
A. 顶点坐标为$(-10,0)$
B. 图象开口向下
C. 函数有最大值
D. 当$x ＜ -10$时,$y$随$x$的增大而减小

答案：

(1)直线$x = 5$

(2)D