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The following equation is: y = the absolute value of X, the absolute value of y = x, 5x & # 178; - y = 0, X & # 178; - Y & # 178; = 0. Where the number of functions where y is x is () A.0 B.1 C.2 D.4
Y is a function of X, which is reflected in: for a certain x, only unique y corresponds to it
The absolute value of y = x and 5x & # 178; - y = 0 satisfy the condition. For the absolute value of y = x, if x = 1, y = 1 or - 1, it does not meet the condition. For X & # 178; - Y & # 178; = 0, if x = 1, y = 1 or - 1, it does not meet the condition
Define f (a, b) = {a, a > b, b > a} then y = the maximum value of F (x, - X & # 178; + 3x)
Let x = - X & # 178; + 3x
Solution
x=0,x=2
So if x > 2 or x-x & # 178; + 3x, then y = f (x, - X & # 178; + 3x) = X
So the maximum is 2
When 0
Find the minimum value of the function y = x & # 178; - 2x-a in the interval [T, t + 1]
F (x) = (x-1) ^ 2-a-1
Is a parabola with vertex (1, - A-1) opening upward
When t + 1 is less than or equal to 1, G (T) = f (T) = T ^ 2-2t-a when t = 1
When t
The definition field of quadratic function is r, what is △?
I don't understand why △ is greater than 0 when the definition field is r? Doesn't △ refer to the number of intersections between the function and the X axis? Doesn't the definition field refer to the value range of the function x? How can I feel that no matter how many intersections there are between the function and the image, the definition field is r?
Your understanding is right. When the domain is r, △ has nothing to do with 0
Given that the domain of quadratic function is r, why is △ less than 0?
Title: y = 2kx-1 / KX square + 4kx + 3, define the field as R, find the value range of real number K
When k is not equal to 0, why should △ be less than 0? I can't understand what the teacher said
Who can give me a detailed explanation, thank you
Because the domain is r, there is no x such that the denominator KX squared + 4kx + 3 = 0
If △ is not less than 0, the equation KX square + 4kx + 3 = 0 has a solution
The image of function y = 12x − 4 is given, and the questions are answered according to the image: (1) when x takes what value, Y > 0; (2) when - 1 ≤ x ≤ 2, the value range of Y is obtained
(1) Drawing can be: from the image, when x ＞ 8, y ＞ 0; (2) first express the unknown x with y, and then calculate the range of Y according to the value range of X. from y = 12x-4 to x = 2Y + 8, that is - 1 ≤ 2Y + 8 ≤ 2, the solution is - 92 ≤ y ≤ - 3. Answer: the value range of Y is - 92 ≤ y ≤ - 3
Draw the image of function y = half x-4, 1 when x takes what value, Y > 0 when - 1 ≤ x ≤ 2, find the value range of Y
When x equals - 1, y = - 4.5, so when - 1 ≤ x, - 4.5 ≤ y
Similarly, y ≤ - 3, so - 4.5 ≤ y ≤ - 3
The image of quadratic function y = x2-4x + 3 does not pass through many quadrants
Can you list how to calculate the axis of symmetry, the vertices? Because it has been discarded for several years, now I don't understand it at all!
Axis of symmetry x = - B / 2A = 4 / 2 = 2
Vertex y = (4ac-b & # 178;) / (4a) = (12-16) / 4 = - 1
So, vertex (2, - 1)
Intersection with Y-axis (0,3)
Obviously, drawing does not go through the third quadrant
For the graph of a linear function y = - 3 / 4x + 2, y passes through the () quadrant with the increase of X
The image passes through the first, second and fourth quadrants. With the increase of X, y passes through the fourth quadrant. When x = 3 / 8, it is a symmetrical image
If the solution of the equations {4x-y = 1, y = 2x + 3} is, then the coordinates of the intersection of the first-order function y = 4x-1 and the image are
The solution is (x = 2, y = 7)
Is it the intersection of y = 4x-1 and 4x-y = 1?
（2,7）
RELATED INFORMATIONS
- 1. (2) given y = f (x) = x & # 178; - 2x + 3, when x ∈ [T, t + 1], find the maximum and minimum function values of the function
- 2. The known function f (x) = 2x-1 g (x) = {X & # 178;, X ≥ 0} (piecewise function) - 1, x < 0 Why do we need to find the norm of X again to find the analytic expression of G [f (x)] g（x）={x²，x≥0 } -1，x＜0
- 3. If f (x) = 4, what is the value of X
- 4. 1. Construct a system of quadratic equations of two variables by image method, and make its corresponding linear function pass through (- 3, - 2), (- 1,6) two points 2. Given that the image intersection of the first-order function y = 2x-a and y = 3x + B and a point outside the origin on the x-axis, what is a / (a + b) equal to?
- 5. Bivariate quadratic system of application problems, listed bivariate linear equations and solutions 1. There are three rooms and two rooms in a hotel. The three rooms are 25 yuan per person per day, and the two rooms are 35 yuan per person per day. It costs 1510 yuan for a tour group of 50 people to stay in the hotel. How many rooms do you rent for each of the two rooms? 2. A stadium runway 400 meters, a, B reverse line, meet once every 30 seconds, the same direction, every 80 seconds B catch up with a, a, B speed respectively?
- 6. One variable quadratic equation test questions 1. It is known that a is a root of the univariate quadratic equation x2-3x + M = 0 about X, and - A is the univariate quadratic equation x2 + 3x-m = 0 about X. try to find the value of A 2. If we know that the equation (K2 + 2) x2 + (5-K) x = 1-3kx2 is a univariate quadratic equation about X, can you find the value of K? 3 (x2 + 3x + 4) (x2 + 3x + 5) = 6 It is known that m.n is the two real number roots of the equation X2 - (P-2) x + 1 = 0 about X. find the value of the algebraic formula (M2 + MP + 1) (N2 + NP + 1)
- 7. Exercises and answers for solving quadratic equation of one variable x*x+2x=36
- 8. A problem about quadratic equation of one variable If the equation x & sup2; + MX + 1 = 0 and the equation x & sup2; - x-m = 0 have only one real root, the value of M is obtained [note]: & sup2; means square; it's better to do it by formula method. I don't know the right answer-|||
- 9. On quadratic equation of one variable from the point of view of function
- 10. Quadratic function and quadratic equation of one variable! 1. The number of intersections of quadratic function y = 2 (x + 2) (x-1) and x-axis is -- and the coordinates of the intersections are—— 2. When m -- the parabola y = x ^ - x + m has an intersection with the X axis 3. If the quadratic function y = 3x ^ + 3x-6 has an intersection with the x-axis, and the coordinates of the intersection are -, then we can see that the quadratic equation 3x ^ + 3x-6 = 0 has a solution, that is, the solution is - 4. Given quadratic function y = 2x ^ - (M + 1) x + (m-1) (1) It is proved that no matter what the value of M is, the image of function y always has an intersection with X axis, and it is pointed out that when the value of M is, there is only one intersection with X axis 5. It is known that the parabola y = ax ^ + 4x + 3 has only one intersection point a with the X axis, and the intersection point with the Y axis is B. try to find the coordinates of a and B and the length of the line ab 6. Given that the image of quadratic function y = x ^ - 2x-3 intersects with X axis at points a and B, there is a point C on the parabola above X axis, and the area of triangle ABC is 10, the coordinate of point C is obtained One two three questions only need to give the result directly. Four five six three questions need the concrete process
- 11. Function y = (SiNx + 1) / (1 + X & # 178;) why is this a bounded function? How to judge
- 12. Function y = 3x + 2 image and y = 1 / X image intersection set
- 13. The set of intersection points between the image of function y = x + 1 and the image of function y = x * 2 + a (constant a ∈ R) is represented by enumeration Quick return
- 14. If the square-2x-1 = 0 of the quadratic equation NX has no real root, then the image of the linear function y = (n + 1) x-n does not pass through the image line
- 15. Given that the parabola y = ax square + BX + C satisfies the following conditions, find the analytic expression of the function (1) Image passing point (- 1. - 6) (1, - 2) (2.3) (2) Vertex coordinate (- 1. - 1) and passing point (0. - 3) (3) Intersection with X-axis (- 2.0) (4,0) and passing through point (1, - 9 / 2)
- 16. It is known that a parabola is obtained by translating the parabola y = ax & # 178; upward by three unit lengths, and the parabola passes through the point (1,0), and its function analytic formula is obtained
- 17. If the image of quadratic function f (x) passes through point (0,0), and f (x +!) = f (x) + X + 1, find the analytic expression of F (x)
- 18. If n people attend the seminar and every two people shake hands once, the analytic formula between the number of handshakes y and the number of participants n is____ It is____ function
- 19. According to his experience, a retailer estimated that (3000-100x) toy dinosaurs could be sold if they were sold at the price of X yuan per dinosaur. It is known that the cost of toy dinosaurs is 20 yuan 1. Write the formula of Y represented by X 2. Judge how many yuan each toy dinosaur sells, y is the biggest
- 20. Find the analytic expression of quadratic function It is known that two points a (12,82) B (14,81) B are the vertices of the function, and the solution set with F (x) greater than 80 is obtained Wrong! Point a is the vertex.