It is known that the image of a first-order function passes through a point (- 2,0), and the area of the triangle formed by a straight line and two coordinate axes is 6

It is known that the image of a first-order function passes through a point (- 2,0), and the area of the triangle formed by a straight line and two coordinate axes is 6

Let y = KX + B, K ≠ 0. Then the intersection point with y axis is (0, b) s △ = 12 × | - 2 × | - B | = 6, then | - B | = 6, | - B = ± 6. When B = 6, the function is y = KX + 6, ∵ the image of the function passes through the point (- 2, 0), then 0 = - 2K + 6, and K = 3. The analytic expression of the first-order function is y = 3x + 6; when B = - 6

It is proved by definition that the function y = 1 - 1 / X is an increasing function on (- ∞, 0)

It is proved that if X1 and X2 belong to (- ∞, 0) and x1 ＜ X2, then f (x1) - f (x2) = (1-1 / x1) - (1-1 / x2) = 1 / x2-1 / X1 = (x1-x2) / x1x2 is determined by x1, and if x2 belongs to (- ∞, 0), then x1x2 ＞ 0, and if x1 ＜ X2, then x1-x2 ＜ 0, namely (x1-x2) / x1x2 ＜ 0, namely f (x1) - f (x2) ＜ 0, namely function y = 1 - 1 / X

Why is definite integral of odd function 0 in symmetric interval? Is there odd function * even function = odd function?

Above the x-axis is positive, below the x-axis is negative
The odd function is symmetric with respect to the origin, so the area of the two blocks in the symmetric interval with respect to the origin is equal, the sign is opposite, and the sum is 0
An odd function multiplied by an even function results in an odd function

Why is the integral of an odd function on a symmetric interval zero? Are not all definite integrals positive?

The positive part and the negative part of an odd function in a symmetric interval are exactly equal, so the definite integral is 0