If the function y = (1 / 3) ax ^ 3 - (1 / 2) ax ^ 2 (a is not equal to 0) is an increasing function in the interval (0,1), then the value range of real number a is

If the function y = (1 / 3) ax ^ 3 - (1 / 2) ax ^ 2 (a is not equal to 0) is an increasing function in the interval (0,1), then the value range of real number a is

Y '= ax ^ 2-ax, y is an increasing function on (0,1)
That is, on (0,1), y '> - 0
ax^2-ax>0
a(x^2-x)>0
Due to 0

Finding the maximum value of function y = ax + 1 (a is not equal to 0) on [0,2]

a> 0, y = ax + 1 is an increasing function, ymax = a * 2 + 1 = 2A + 1, Ymin = a * 0 + 1 = 1
a

Given the intersection of the function y = ax square (a is not equal to 0) and the straight line y = 2x-3 at the point a (1, b), find the values of (1) a and B (2) the vertex coordinates and the axis of symmetry of the parabola y = ax square (3) when x takes what value, y in the quadratic function y = ax square increases with the increase of X? (4) if you want to find the area of the triangle formed by the two intersections and the vertices of the parabola y = ax square and the straight line y = - 2, do not answer the first three questions,

(1) Intersection a (1, b)
y=2x-3
b=2*1-3=-1
y=ax^2
-1=a*1^2
a=-1
(2) y=-x^2
Vertex: O (0,0)
Axis of symmetry: x = 0
(3) Opening down, X

The image of the function y = ax ^ 2 + BX + C (a is not equal to zero) is a parabola. Find its focus and directrix

Guide line: y = - 1 / 4 + (4ac-b ^ 2) / 4A
Focus: (- B / 2a, 1 / 4 + (4ac-b ^ 2) / 4A)
The formula of the equation in the title can get the translation relationship between the equation and y = x ^ 2, and then the translation relationship between the guide line of y = x ^ 2 and the focus

Given that the set P = (1 / 2,2) and the function y = log2 (AX square - 2x + 2), the domain of definition is Q. if the intersection of P and Q is not equal to the empty set, the value range of a is obtained

The domain of y = log2 (AX squared - 2x + 2) is Q, if P intersection q is not equal to an empty set
Explain X in (1 / 2,2)
Ax squared - 2x + 2 > = 1
ax^2-2x+1>=0
1> When a > 0,2 / A2, that is a > 4 or 0

Let f (x) = X3 + AX2 + (a + 6) x + 1 have both maxima and minima, then the value range of real number a ()
A. A ＞ 6 or a ＜ - 3B. - 3 ＜ a ＜ 6C. A ≥ 6 or a ≤ - 3D. - 3 ≤ a ≤ 6

∵ f (x) = X3 + AX2 + (a + 6) x + 1 ∵ f '(x) = 3x2 + 2aX + (a + 6) ∵ function f (x) = X3 + AX2 + (a + 6) x + 1 has both maximum and minimum ∵ = (2a) 2-4 × 3 × (a + 6) > 0 ∵ a > 6 or a < - 3, so select a

Given the function f (x) = - X & # 178; + 2ax-3a, when the maximum value of F (x) in the interval [1,2] is 4
I'm not good at finding the value of real number a

f(x)=-(x-a)^2-3a+a^2
Discussion a:
When 1=

Find the necessary and sufficient condition of y-axis-to-axis for the image of function y = ax ^ 2 + BX + C

First of all, he said that the premise is a function, not a quadratic function, so a can be equal to 0
If a = 0, B = 0, it holds
When a is not equal to 0, - (B / 2a) = 0, B = 0 because it is symmetric about the y-axis
So the necessary and sufficient condition is b = 0

If and only if the image of quadratic function y = ax ^ 2 + BX + C is above the x-axis

If the image of quadratic function y = AX2 + BX + C is above the x-axis, b2-4ac ＜ 0, a ＞ 0

If the image of function f (x) = a ^ x + B + 1 (a > 0 and a is not equal to 1) passes through the second, third and fourth quadrants, there must be:
A. 01 and b > 0
C.0

Because in quadrant 2, 3 and 4, the function decreases by 0