描绘 y=1+[36x/(x+3)²]的图形
解:定义域:x≠-3.
x→∞limy=x→∞lim[1+36x/(x+3)²]=1; 因此有水平渐近线y=1;
x→-3-limy=-∞; x→-3+limy=-∞; 因此有垂直渐近线x=-3;
令y'=[36(x+3)²-72x(x+3)]/(x+3)^4=[36(x+3)-72x]/(x+3)³=36(3-x)/(x+3)³=0
3-x=0,故得驻点x=3;
x<3时y'>0;x>3时y'<0;∴x=3时极大点,极大值f(x)=f(3)=1+3=4;
![](https://iknow-pic.cdn.bcebos.com/9d82d158ccbf6c81d7cd53f3b53eb13533fa4036?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)