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tangent f(x)=sqrt(x^2+13),\at x=6

解答

切线

f (x) = x^{2} + 13, at x = 6

解答

y = \frac{6}{7} x + \frac{13}{7}

求解步骤

找到切点:

(6, 7)

计算

f (x) = x^{2} + 13

的斜率:

\frac{df ( x )}{d x} = \frac{x}{x ^{2} + 13}

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{6}{7}

找到斜率 m=

\frac{6}{7}

且穿过

(6, 7)

的直线:

y = \frac{6}{7} x + \frac{13}{7}

y = \frac{6}{7} x + \frac{13}{7}

作图

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