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tangent f(x)=4x^2+7x

解答

切线

f (x) = 4 x^{2} + 7 x

解答

y = (8 a_{0} + 7) x - 4 a_{0}^{2}

求解步骤

计算在一般点

x = a_{0}

的切线

找到切点:

(a_{0}, 4 a_{0}^{2} + 7 a_{0})

计算

f (x) = 4 x^{2} + 7 x

的斜率:

\frac{df ( x )}{d x} = 8 x + 7

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 = 8 a_{0} + 7

找到斜率 m=

8 a_{0} + 7

且穿过

(a_{0}, 4 a_{0}^{2} + 7 a_{0})

的直线:

y = (8 a_{0} + 7) x - 4 a_{0}^{2}

y = (8 a_{0} + 7) x - 4 a_{0}^{2}

作图

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