Y = ax ^ 2 + BX is an even function on the definition {a-1,2a}, and the value of a + B is obtained

Y = ax ^ 2 + BX is an even function on the definition {a-1,2a}, and the value of a + B is obtained

Y = ax ^ 2 + BX is even function on the definition {a-1,2a}
So the domain {a-1,2a} is symmetric about the origin, that is, A-1 + 2A = 0, so a = 1 / 3
Let y = f (x) = ax & sup2; + BX, then f (- x) = a (- x) & sup2; - BX = f (x) = ax & sup2; + BX
That is, a (- x) & sup2; - BX = ax & sup2; + BX, so B = 0
a+b=1/3+0=1/3
When solving the problem of even function, we only need to grasp a few points, the domain of definition is symmetric about the origin, f (- x) = f (x)

Given that the function f (x) = ax square + BX + B is even and the domain of definition is [a-1,2a], then a =?, B =?

Because the domain of even function is symmetric about the origin
So A-1 + 2A = 0
So a = 1 / 3
And the y-axis is the symmetry axis of the function
So - B / 2A = 0
So B = 0
So a = 1 / 3, B = 0

It is known that FX is equal to the square of AX + BX + 3A + B, which is an even function defined on [A-4, a]

F (x) = ax & # 178; + BX + 3A + B are even functions
Then the domain of definition is symmetric about the origin, that is, 4-A = a, and the solution is a = 2
f(-x)=2x²-bx+6+b
f(x)=f(-x)
So B = - B; the solution is b = 0
f(x)=2x²+6
Function range [6,14]

6. The image of function y = LG (1 + 2 / x minus 1) about ()
(A) X-axis symmetry (b) Y-axis symmetry (c) origin symmetry (d) straight line y = x-symmetry
Why?

It's y = LG [2 / (x + 1) - 1]
y=lg[(1-x)/(1+x)]
f(-x)=lg[(1+x)/(1-x)]=-lg[(1-x)/(1+x)]=-f（x）
It is an odd function
(c) origin symmetry

Given a function y = 2x + 1, the intersection coordinate of this function and Y axis is______ ．

∵ when x = 0, y = 1. The coordinates of the intersection of ∵ and Y axis are (0, 1)

It is known that the ordinate of the intersection point of the inverse scale function y equals to K / X and the first-order function y equals to 2x plus K is negative. What is k equal to

y=k/x (1)
y=2x+k (2)
Substituting y = - 4 into (1) yields
x=-k/4
Substitute it into (2) to get
-4=2*(-k/4)+k
The solution is k = - 8

Take a point P on the image of the function y = - 3x, pass through the point P and make the vertical X axis of PA. if the abscissa of the point P is known to be - 2, then the area of the triangle POA (o is the origin of the coordinate) is______ ．

When x = - 2, y = - 3 × (- 2) = 6, the coordinates of point P are (- 2,6), PA = 6, OA = | - 2 | = 2, s △ POA = 12 × PA × OA = 12 × 6 × 2 = 6

In order to make the image of function y = {2m-3} x + {3N + 1} pass through the positive half axis of X and Y axes, the values of M and n should be?
The previous one is the process and answer of solving the problem in parentheses

2m-3＜0
3n+1＞0
That is, m < 3 / 2
n＞-1/3

In order to make the image of function y = (2m-3) x + (3N + 1) pass through the positive half axis of X and Y axes, the values of M and n should be ()
A. m＞32，n＞-13B. m＞3，n＞-3C. m＜32，n＜-13D. m＜32，n＞-13

∵ the image of the function y = (2m-3) x + (3N + 1) passes through the positive half axis of X and Y axes, and the result is 2m − 3 ＜ 03n + 1 ＞ 0, and the solution is m ＜ 32, n ＞ - 13

For a linear function y = (M + 4) x + 2m-1, if y increases with the increase of X, and the intersection of its image and Y axis is below X axis, try to find the value range of M

∵ the first-order function y increases with the increase of X, ∵ k = m + 4 ＞ 0, that is, m ＞ - 4. The intersection of the image of the first-order function and the Y axis is below the X axis, ∵ B = 2m-1 ＜ 0, that is, m ＜ 12. The value range of M is - 4 ＜ m ＜ 12