知识点地图 / 进阶数学
Discriminant, Tangency & Number of Roots
判别式・相切与根的个数
把交点问题化成一元二次方程。先判断判别式的符号，再处理参数范围、整数边界和重根坐标。
这部分需要会什么？
- 用判别式判断实根个数
- 将直线与抛物线的交点问题转为二次方程
- 结合参数限制求相切或无交点条件
核心概念与公式
对真正的二次方程 \(ax^2+bx+c=0\)（\(a\ne0\)），判别式 \(\Delta=b^2-4ac\)。\(\Delta>0\) 有两个不同实根，\(\Delta=0\) 有一个不同实根（重根），\(\Delta<0\) 无实根。
求直线与抛物线交点时，先联立，整理成关于同一个变量的二次方程。相切对应重根，因而判别式为零。若含参数使二次项系数为零，方程可能变成一次方程，必须另外判断。
做题时怎样使用
- 先整理标准式，明确带符号的 \(a,b,c\)
- 把“两个交点、相切、无实根”翻译成判别式的条件
- 解参数条件后检查 \(a\ne0\)，再应用正数、整数等题目限制
容易混淆的地方
判断两条曲线的交点，必须用联立后方程的判别式，不能只看抛物线自身与横轴的交点。
用一个例题串起方法
知识点教学例题直线 \(y=2x+k\) 与抛物线 \(y=x^2+4\) 相切，求 \(k\)。
展开解题思路 ＋
联立得 \(x^2-2x+4-k=0\)。相切要求 \(\Delta=4-4(4-k)=0\)，所以 \(k=3\)。此时方程为 \((x-1)^2=0\)，交点为 \((1,5)\)。
本知识点题库
先独立作答，再查看答案与解析。
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