1. If the line y = - 2x + 3 intersects the parabola y = ax at two points a and B, and the coordinates of point a are (- 3,9), what are the opening direction, symmetry axis and vertex coordinates? 2. Given that the area of a rectangle is 6, find the functional relation between its length y and width x, and draw the image of this function in rectangular coordinate system 3. The quadratic function y = x-2x + 1 is known (1) Find the vertex a of this function image and the coordinates of its intersection B with y axis (2) Find the coordinates of the intersection points c and D of the function image and X axis; (3) Find ABC of s Triangle 3. The quadratic function y = 1 / 2x & sup2; - 2x + 1 is known (1) Find the coordinates of the vertex a of the function image and its intersection B with the y-axis. (2) Find the coordinates of the intersection points c and D of the function image and X axis; (3) Find ABC of s Triangle

1. If the line y = - 2x + 3 intersects the parabola y = ax at two points a and B, and the coordinates of point a are (- 3,9), what are the opening direction, symmetry axis and vertex coordinates? 2. Given that the area of a rectangle is 6, find the functional relation between its length y and width x, and draw the image of this function in rectangular coordinate system 3. The quadratic function y = x-2x + 1 is known (1) Find the vertex a of this function image and the coordinates of its intersection B with y axis (2) Find the coordinates of the intersection points c and D of the function image and X axis; (3) Find ABC of s Triangle 3. The quadratic function y = 1 / 2x & sup2; - 2x + 1 is known (1) Find the coordinates of the vertex a of the function image and its intersection B with the y-axis. (2) Find the coordinates of the intersection points c and D of the function image and X axis; (3) Find ABC of s Triangle

1. "Parabola y = ax & sup2;"
The point a (- 3,9) is on the parabola y = ax & sup2
9 = a (- 3) & sup2;, that is, a = 1
The parabola is y = x & sup2;,
The opening is upward, the axis of symmetry is x = 0, and the vertex coordinates are (0,0);
2、∵xy=6
∴y=6/x,
The functional relation between its length y and width x is y = 6 / X (x > 0),
Let x = 2, y = 3
Let x = 3, y = 2
The image of the function can be obtained by taking points (2,3), (3,2) in the rectangular coordinate system;
3. (1) ∵ quadratic function y = 1 / 2x & sup2; - 2x + 1 = 1 / 2 (X-2) & sup2; - 1,
The vertex a (2,1) of the graph of this function
∵ its intersection with the Y axis is B,
Let x = 0, then y = 1
The coordinate (0,1) of the intersection point B between it and y-axis
(2) The intersection points of this function image and X axis are C and D,
Let y = 0, then 1 / 2 (X-2) & sup2; - 1 = 0
∴x=2±√2,
The coordinates of the intersection point C and D of this function image and X axis are (2 + √ 2,0), (2 - √ 2,0) respectively
Or (2 - √ 2,0), (2 + √ 2,0)
（3）∵A（2,1）,B（0,1）
{ab ‖ x-axis, ab = 2
When the coordinate of point C is (2 + √ 2,0), CD ⊥ AB is made at d through point C
Then CD = 1
∴S△ABC=1/2 * AB * CD=1/2×2×1=1,
When the coordinate of point C is (2 - √ 2,0), CE ⊥ AB is made at e through point C
Then CE = 1
∴S△ABC=1/2 * AB * CE=1/2×2×1=1,
So the area of triangle ABC is 1
When solving area,
Because ab ∥ x-axis, the area of any point on x-axis and the triangle formed by a and B is 1,
The reason is: the height is 1, the bottom AB = 2

How to solve X in cubic function with the knowledge of high school mathematics?
One method is to solve the factor But I'm weak in this aspect! I haven't even reacted to cross multiplication up to now How to deal with cubic function solution X in college entrance examination
For example, the answer of 2x ^ 3-3x ^ 2 + 1 = 0 is x = 1 or - 1 / 2 Cubic function is to reduce the power of the second, to find the detailed process of this problem!
It should not be able to seek derivation! I have sought guidance, but I can't calculate X!

Pro, generally test is special~_~
The knowledge point used is the square difference formula and the cubic expansion~_~
a²-b²=(a+b)(a-b) a³-b³=(a-b)(a²+ab+b²)
Do your question to give you some inspiration
Original formula = 2 (X & # 179; - X & # 178;) + 1-x & # 178; = 2x & # 178; (x-1) - (x + 1) (x-1) = (x-1) (2x & # 178; - x-1) = (x-1) (2x + 1) (x-1) = (x-1) &# 178; (2x + 1)
So the answer is x = 1 or x = - 1 / 2
[the test is not too difficult. It depends on whether you are careful or not. In fact, what you test is factoring]
Good luck~_~

As shown in the figure, the vertex a of RT △ ABO is the intersection of hyperbola y = KX and straight line y = - X - (K + 1) in the second quadrant. Ab ⊥ X axis is in B, and s △ ABO = 32. (1) find the analytic expressions of the two functions; (2) find the coordinates of two intersections A and C of straight line and hyperbola and the area of △ AOC

(1) Let the coordinates of point a be (x, y), and X ＜ 0, y ＞ 0, then s △ ABO = 12 · | Bo · · · · · · · · BA · = 12 · (- x) · y = 32, ∧ xy = - 3, and ∧ y = kx, that is, xy = k, ∧ k = - 3. The analytic expressions of the two functions obtained are y = - 3x, y = - x + 2 respectively; (2) by y = - x + 2, let x = 0, y = 2 1 = - 1y1 = 3, X2 = 3y2 = - 1, intersection a is (- 1,3), C is (3, - 1), s △ AOC = s △ ODA + s △ ODC = 12od · (| x1 | + | x2 |) = 12 × 2 × (3 + 1) = 4

Challenge your knowledge of mathematical functions
The intersection points P (- 2,0), q (1, m) of this function and inverse scale function are known
1. Find the relationship between the two functions
2. Draw the image of these two functions in the same coordinate system. According to the function answer, when x is the value, the value of the primary function is greater than that of the inverse proportion function
Sorry, P is (- 2, 1)

Because the inverse proportion function passes the point P (- 2,1), so: in y = K / x, k = - 2 * 1 = - 2, that is, y = - 2 / X replaces Q (1, m) into y = - 2 / x, M = - 2. So: the solution of the first-order function y = KX + B - 2K + B = 1K + B = - 2 is k = - 1, B = - 1, that is, y = - X-12

F (x) satisfies the condition f (x + 2) = 1F (x) for any real number X. if f (1) = - 5, then f (f (5)) = ()
A. -5B. −15C. 15D. 5

∵ f (x + 2) = 1F (x) ∵ f (x + 2 + 2) = 1F (x + 2) = f (x) ∵ f (x) is a function with a period of 4 ∵ f (5) = f (1 + 4) = f (1) = - 5F (f (5)) = f (- 5) = f (- 5 + 4) = f (- 1) and ∵ f (- 1) = 1F (− 1 + 2) = 1F (1) = - 15 ∵ f (f (5)) = - 15

It is known that the function f (x) is an odd function defined on R. when x > 0, f (x) = x (x + 1), the analytic expression of the function is obtained

Set x0
So f (- x) = - x (- x + 1) = x ^ 2-x
F (x) = - f (- x) for odd functions
So, x0)
f(x)=0,(x=0)
f(x)=-x^2+x,(x

If the solution set of inequality (1 + K2) x ≤ K4 + 4 about X is m, then for any real constant K, there is always ()
A. 2∈M，0∈MB. 2∉M，0∉MC. 2∈M，0∉MD. 2∉M，0∈M

Method 1: substituting x = 2 and x = 0 into the inequality to judge whether the solution set of the inequality about K is R or not; method 2: finding out the solution set of the inequality: (1 + K2) x ≤ K4 + 4 {x ≤ K4 + 4k2 + 1 = (K2 + 1) + 5k2 + 1 − 2 {x ≤ [(K2 + 1) + 5k2 + 1 − 2] min = 25 − 2; so select a

1. F (x) is a function of degree, and if 3f (x + 1) - f (x) = 2x + 9, f (x)
2. Given that f (x) is a quadratic function and satisfies f (0) = 0, f (x + 1) - f (x) = 2x, find the analytic expression of F (x)
3. Given f (x + 1) = x & # 178; + 4x + 1, find the analytic expression of F (x)

1. F (x) is a function of degree, and if 3f (x + 1) - f (x) = 2x + 9, f (x)
∵ f (x) is a function of degree, Let f (x) = KX + B, then we have the following equation according to the known conditions:
3f(x+1)-f(x)=3[k(x+1)+b]-(kx+b)=3kx+3k+3b-kx-b=2kx+3k+2b=2x+9,
So 2K = 2, k = 1; 3K + 2B = 3 + 2B = 9, so B = 3
∴f(x)=x+3.
2. Given that f (x) is a quadratic function and satisfies f (0) = 0, f (x + 1) - f (x) = 2x, find the analytic expression of F (x)
∵ f (x) is a quadratic function, Let f (x) = ax & # 178; + BX + C, then we have the following equation:
F (0) = C = 0, that is, C = 0;
F (x + 1) - f (x) = a (x + 1) ² + B (x + 1) - ax & #178; - BX = 2aX + A + B = 2x, so 2A = 2, that is, a = 1; a + B = 1 + B = 0, so B = - 1
So the analytic formula is: F (x) = x & # 178; - X
3. Given f (x + 1) = x & # 178; + 4x + 1, find the analytic expression of F (x)
Solution 1: F (x + 1) = x & # 178; + 4x + 1 = (x + 1) &# 178; + 2 (x + 1) - 2; [through the identity edge, make the independent variable x in the original expression become (x + 1)]
If x + 1 in the above formula is replaced by X, the analytic formula is f (x) = x & # 178; + 2x-2
Solution 2: let x + 1 = u, then x = U-1, and substitute into the original formula to get f (U) = (U-1) & # 178; + 4 (U-1) + 1 = u & # 178; + 2u-2
If u is replaced by X, the analytic formula is: F (x) = x & # 178; + 2x-2

2 ^ (2 + log small 2 big 5)

2 + log small 2 big 5 = log small 2 big 4 + log small 2 big 5
=Log small 2 large (4 × 5)
=Log small 2 big 20 (meaning, if a power of 2 is equal to 20, log small 2 big 20 stands for "some")
2 ^ (log small 2 big 20) = 20

Will write one by one
1. The (A & sup2; - 1) n power of function f (x) = 2 is defined as a decreasing function on R, then the value range of real number a is obtained
2. The X + 3 power-2 (a > 0 and a ≠ 0) of function y = 3 is constant over a fixed point____________
3. Let m = the (A & sup2; + 1) power of loga, n = the 2A power of loga, where a > 1, then the size relation of M n
4. If f (x) is a power function and f (4) is greater than f (2) = 3, then the value of F (2 / 1)
5. Let a contain {- 112 / 13}, then the domain of definition of a power of function y = x is R and all a values of odd function

1、 -1