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normal f(x)=sqrt(x^2+16),\at x=3

解答

法线

f (x) = x^{2} + 16, at x = 3

解答

f (x) = - \frac{5}{3} x + 10

求解步骤

找到切点:

(3, 5)

计算

f (x) = x^{2} + 16

的斜率:

\frac{df ( x )}{d x} = \frac{x}{x ^{2} + 16}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{3}{5}

计算垂线的斜率:

m_{p} = - \frac{5}{3}

找到斜率 m=

- \frac{5}{3}

且穿过

(3, 5)

的直线:

f (x) = - \frac{5}{3} x + 10

f (x) = - \frac{5}{3} x + 10

作图

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