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tangent ax^2+bx+c

解答

切线

a x^{2} + b x + c

解答

y = (2 a a_{3} + b) x - a a_{3}^{2} + c

求解步骤

计算在一般点

x = a_{3}

的切线

找到切点:

(a_{3}, a a_{3}^{2} + b a_{3} + c)

计算

y = a x^{2} + b x + c

的斜率:

\frac{d y}{d x} = 2 a x + b

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 = 2 a a_{3} + b

找到斜率 m=

2 a a_{3} + b

且穿过

(a_{3}, a a_{3}^{2} + b a_{3} + c)

的直线:

y = (2 a a_{3} + b) x - a a_{3}^{2} + c

y = (2 a a_{3} + b) x - a a_{3}^{2} + c

作图

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