Given that the vertex coordinates of the quadratic function y = (a + 2) x & # 178; + A & # 178; - 1 are (0,3), then a=_____
Substituting (0,3) into y = (a + 2) x & # 178; + A & # 178; - 1, we get 3 = A & # 178; - 1, a & # 178; = 4, we get a = 2 or - 2, because when a = - 2, y = (a + 2) x & # 178; + A & # 178; - 1 = 3, which is not satisfied, so we know that the vertex coordinate of quadratic function y = (a + 2) x & # 178; + A & # 178; - 1 is (0,3), then a=_ 2__ ...
RELATED INFORMATIONS
- 1. What is the vertex coordinate of quadratic function y = (x-1) - 2?
- 2. The vertex coordinates of the image of quadratic function y = x & # 178; + 1 are
- 3. It is known that the quadratic function y = x square - (m square + 4) x-2m square-12 1. Verification: no matter what real number m takes, the parabola passes through a certain point, and the coordinates of the fixed point are obtained
- 4. Given that the image of the function f (x) defined on R is symmetric with respect to the point (- 3 / 4,0) and satisfies f (x) = - f (x + 3 / 2). F (- 1) = 1. F (o) = - 2, then the value of f (1) + F (2) + F (3) +... + F (2008) is () A.-2 B.-1 C.0 D.1
- 5. It is known that the image of the function f (x) defined on R is symmetric with respect to the point (- 3 / 4,0), satisfying f (x) = - f (x + 3 / 2), f (- 1) = 1, f (0) = - 2 Then the value of F (1) + F (2) +... + F (2008) is Please indicate the process
- 6. It is known that the image of the function f (x) defined on R is symmetric with respect to the point (- 3 / 4,0) and satisfies f (x) = - f (x + 3 / 2). F (- 1) = 1. F (o) = - 2, Then the value of F (1) + F (2) + F (3) +... + F (2008) is () A.-2 B.-1 C.0 D.1
- 7. If the domain of F (x) is symmetric about the origin, then f (x) times f (- x) is an even function How to prove it
- 8. The image of the function FX defined on R is symmetric with respect to the point a (a, b) B (C, b), and the period of the function is calculated
- 9. As shown in the figure, the image of the first-order function y equal to minus 2x plus B and the image of the second-order function y equal to minus xsquare plus 3x plus C pass through the origin (1) If the line y = KX + m and the line y = negative 2x + B are parallel to the Y axis and intersect at point a and pass through the vertex P of the parabola to find the expression of the line y = KX + m, and (3) find the area of the triangle apo
- 10. It is known that the image of the first-order function y = - 1 / 2x + 4 intersects with the x-axis and y-axis respectively at a, B, and the edge AC = 5 of the trapezoid aobc (o is the origin) (1) Find the coordinate (2) of point C if points a and C are in the linear function y = KX + B (k, B are constants, and K
- 11. It is known that the quadratic function y = x & # 178; + (2k-1) x + K & # 178; - 1 of X, if the sum of two squares of the quadratic equation x & # 178; + (2k-1) x + K & # 178; - 1 = 0 of X is 9, find the value of K and the vertex coordinates of the parabola, and find the correct vertex coordinates!
- 12. The minimum value of quadratic function y = 15 (x-1) 178; is
- 13. Quadratic function y = a (x-1) B has the minimum value - 1, what is a + B The quadratic function y = a (x-1) 178; + B has a minimum value of - 1, what is a + B
- 14. Given that the minimum value of quadratic function y = x2 + (2a + 1) x + A2-1 is 0, then the value of a is () A. 34B. -34C. 54D. -54
- 15. It is known that the definition domain of quadratic function y = A & # 178; - 2a-1 is (- ∞, - 1) ∪ (3, + ∞) to find the minimum value of Y
- 16. Quadratic function y = ax ^ 2 + BX + C The distance between the two intersection points of quadratic function y = ax ^ 2 + BX + C and X axis is 4 units. If the image is translated 1.5 units along the symmetric axis, then the image passes through the coordinate origin. If the original image is translated 2 units along the symmetric axis, then the image vertex is on the X axis, and the analytical formula of parabola is obtained
- 17. Given the quadratic function y = 3x-5x-2, when what is the value of X, y = 0, Y0
- 18. When B is a value, there is one intersection point between the primary function y = 5x + B and the secondary function y = x ^ 2 + 3x + 5, two intersections and no intersection point?
- 19. If there is an intersection between the image of the first function y = 5x + m and the image of the second function y = x2 + 3x + 5, then the value range of M is
- 20. How to draw the image of quadratic function? I see y = 3x ^ 2 confused How to establish? Told me. I will benefit a lot