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tangent f(x)=x^2+ax+2,\at x=1

解答

切线

f (x) = x^{2} + a x + 2, at x = 1

解答

y = (2 + a) x + 1

求解步骤

找到切点:

(1, a + 3)

计算

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

的斜率:

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

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

找到斜率 m=

2 + a

且穿过

(1, a + 3)

的直线:

y = (2 + a) x + 1

y = (2 + a) x + 1

作图

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