Y=x^3+2x+3求抛物线在点M0(2.11)处的切线方程和法线方程

根据导数的几何意义,函数y ＝ f(x)在点x0处的导数f'(x0)在几何上表示曲线y = f(x)在点M(x0,f(x0))处的切线的斜率,即
f'(x0) = tanα, 其中α是切线的倾角.
由直线的点斜式方程,知曲线y = f(x)在点M(x0,y0)处的切线方程是
y - y0 = f'(x0)(x-x0)
过点M(x0,y0)且与切线垂直的直线叫做曲线y=f(x)在点M处的法线.如果f'(x0)≠0 ,法线的斜率为 -1/f'(x0),从而法线方程为
y - y0 = -1/f'(x0)(x - x0)
对于本题,
y = x^2+2x+3 的导数为
y' = 2x +2
则在点M0(2,11)处的导数为 f'(2) = 2*2 +2 = 6
即过点M0(2,11)的切线斜率为k = 6
则切线方程为
y-11 = 6(x-2)
即 y = 6x - 1
易知法线斜率为 k' = -1/6 = -1/6
所以在点M0(2.11)处的法线方程为
y -11 = -1/6(x-2)
即 x + 6y -68 = 0