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tangent f(2.1)=3x^2-4x+1,(2,0)

解答

切线

f (2.1) = 3 x^{2} - 4 x + 1, (2, 0)

解答

y = 2 (4 + 15) x - 4 (4 + 15) or y = 2 (4 - 15) x - 4 (4 - 15)

求解步骤

(2, 0)

处的切线斜率:

m = \frac{3 x ^{2} - 4 x + 1}{x - 2}

计算

f (2.1) = 3 x^{2} - 4 x + 1

的斜率:

m = 6 x - 4

解

6 x - 4 = \frac{3 x ^{2} - 4 x + 1}{x - 2} : x = \frac{6 + 15}{3}, x = \frac{6 - 15}{3}

求出

x = \frac{6 + 15}{3}

处的切线:

y = 2 (4 + 15) x - 4 (4 + 15)

求出

x = \frac{6 - 15}{3}

处的切线:

y = 2 (4 - 15) x - 4 (4 - 15)

y = 2 (4 + 15) x - 4 (4 + 15) or y = 2 (4 - 15) x - 4 (4 - 15)

作图

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