截距 f(x)=(4x+1)^3(4x^2-4x+1)

解答

截距

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

X 截距 : (- \frac{1}{4}, 0), (\frac{1}{2}, 0), Y 截距 : (0, 1)

求解步骤

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

的

x -

轴截点:

(- \frac{1}{4}, 0), (\frac{1}{2}, 0)

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

的

y -

轴截点:

(0, 1)

X 截距 : (- \frac{1}{4}, 0), (\frac{1}{2}, 0), Y 截距 : (0, 1)

r an g e - x + 3

e x t re m e y = x^{3} + y^{3} - x - y - 2 = 0

in v erse f (x) = \frac{x}{- 8 x + 3}

mi d p o in t (6, 97), (- 4, - 77)

in v erse f (x) = \frac{1}{x - 10}