General formula: y = ax ^ 2 + BX + C (a, B, C are constants, a ≠ 0) vertex formula: y = a (X-H) ^ 2 + k The general formula A, B, C is about how to move the image, and the vertex formula A, h, K is about what to move the image For example, y = ax, square a controls the up and down movement of the image

General formula: y = ax ^ 2 + BX + C (a, B, C are constants, a ≠ 0) vertex formula: y = a (X-H) ^ 2 + k The general formula A, B, C is about how to move the image, and the vertex formula A, h, K is about what to move the image For example, y = ax, square a controls the up and down movement of the image

General formula: y = ax ^ 2 + BX + C (a, B, C are constants, a ≠ 0),
A determines the direction of the opening, - B / 2A determines the sign of the axis of symmetry, and C determines the intercept
In the vertex formula: y = a (X-H) ^ 2 + K,
A determines the opening direction, H determines the axis of symmetry and the abscissa of the vertex, and K determines the ordinate of the vertex

Let f (x) be differentiable at point x = a, and find Lim [f (a) - f (a - △ x)] / △ x △ x → 0
lim[f(a)-f(a-△x)]/△x
=-lim[f(a)-f(a-△x)]/(-△x)
Why is the denominator - △ x
Please give specific reasons,

The definition of derivative is f '(a) = Lim [f (a) - f (a + △ x)] / △ x △ x → 0
Instead of F '(a) = Lim [f (a) - f (a - △ x)] / △ x △ x → 0
Note that there is a plus sign in the middle, not a minus sign

It is proved that if f (x) is continuous at x = 0 and lim (x → 0) (f (x) / x) = 1, then f '(0) = 1

Because LIM (x → 0) (f (x) / x) = 1, X and f (x) are equivalent infinitesimals: F (x). When x tends to 0, f (x) also tends to 0
So: F (0) = 0
f'(0)= lim(x→0) [f(x)-f(0)]/(x-0)
= lim(x→0) f(x)/x
= 1