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tangent 4x^2-2x+3

解答

切线

4 x^{2} - 2 x + 3

解答

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 3

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} - 2 a_{0} + 3)

计算

y = 4 x^{2} - 2 x + 3

的斜率:

\frac{d y}{d x} = 8 x - 2

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} - 2

找到斜率 m=

8 a_{0} - 2

且穿过

(a_{0}, 4 a_{0}^{2} - 2 a_{0} + 3)

的直线:

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 3

y = (8 a_{0} - 2) x - 4 a_{0}^{2} + 3

作图

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