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Since the independent variable x ≠ 0 and the function value y ≠ 0 in the inverse scale function y = K / X (K ≠ 0), its image is at the intersection of X axis and Y axis, that is, the two branches of hyperbola are infinitely close to 2, but never reach the coordinate axis. In addition, the two branches of hyperbola are symmetrical about coordinate 3, and about straight line y = x and straight line y = - x 4
Since the independent variable x ≠ 0 and the function value y ≠ 0 in the inverse scale function y = K / X (K ≠ 0), its image has no intersection with the X axis and Y axis, that is, the two branches of the hyperbola are infinitely close to the coordinate axis 2, but never reach the coordinate axis. In addition, the two branches of the hyperbola are symmetrical about the centroid of the origin of coordinate 3, and about the axis of the straight line y = x and the straight line y = - x 4
The focus of the ellipse is on the x-axis, the length of the short half axis is 4, and the eccentricity is 1 / 2. Try to find its standard equation
c/a=1/2,b=4,a²-b²=c²
The results show that a = 8 √ 3 / 3, B = 4, C = 4 √ 3 / 3 = > A & sup2; = 64 / 3, B & sup2; = 16
Standard equation: 3x & sup2 / 64 + Y & sup2 / 16 = 1
In the function y = - 3 + 2, when the independent variable x satisfies --------, the image is in the first quadrant
Y=-3X+2
The image is in the first quadrant
Then x > 0, Y > 0
There are - 3x + 2 > 0, x > 0
The solution is 0
Find the standard equation of ellipse whose focus is (- 2,0), (2,0) and eccentricity is 1 / 3, and write the vertex coordinates
c=2 e=c/a=2/a=1/3 ∴a=6
∴b²=a²-c²=36-4=32
The equation is X & # 178 / 36 + Y & # 178 / 32 = 1
Vertex (- 6,0) (6,0) (0,4 √ 2) (0, - 4 √ 2)
The value of quadratic function y = - X & # 178; - 6x + 3 decreases with the increase of X, and the value range of independent variable x
y=-x²-6x+3
=-(x^2+6x+9)+3+9
=-(x+3)^2+12
The image on the right side of the symmetry axis X = - 3 decreases with the increase of X
So x ≥ - 3
y=-x²-6x+3
=-(x-3)^2+12
The parabolic opening is downward. So when x > = 3, it's a decreasing function
The image on the right side of the opening downward symmetry axis decreases with the increase of X
So x ≥ - 3
This is obviously a parabola with an opening downward. Using the formula of the axis of symmetry, we find that the axis of symmetry is x = 3, so the value range of X is x > 3
Y = - (x ^ 2 + 6x + 9) + 12 = - (x + 3) ^ 2 + 12, then the symmetry axis of quadratic function is x = - 3 and the opening is drawn downward,
When x belongs to [- 3, + ∞), y decreases with the increase of X
y=-x²-6x+3
∵ is a quadratic function
∴x≠0
Formula y = - (X & # 178; + 6x-3)
y=-【（x+3）²-12】
y=-(x+3)²+12
The value of - 6x + 3 decreases with the increase of X
∴x≠0，x≥-3
1. One vertex coordinate (- 4,0), one focus coordinate (0,3) to solve the elliptic standard equation? 2
Urgent: 1. One vertex coordinate (- 4,0), one focus coordinate (0,3) to solve the elliptic standard equation? 2. The distance between two collimators is 4, the length of the short half axis is 1, to solve the elliptic standard equation?
1. If the elliptic equation is Y & # 178; / A & # 178; + X & # 178; / B & # 178; = 1 (a > b > 0), and B = 4, C = 3, then a & # 178; = B & # 178; + C & # 178; = 25, the ellipse is Y & # 178; / 25 + X & # 178; = 1
2. The distance between the two collimators 2A & # 178 / C = 4, B = 1, a & # 178; = 2, B & # 178; = 1, C & # 178; = 1, the ellipse is X & # 178 / 2 + Y & # 178; = 1 or X & # 178; + Y & # 178 / 2 = 1
If B = 4, C = 3, then a = 5, so the elliptic equation is y ^ 2 / 25 + x ^ 2 / 16 = 1
The distance between the two guide lines is a ^ 2 / C * 2 = 4, and the short half axis is b = 1
Then we can get a ^ 2-B ^ 2 = 2c-1 = C ^ 2, the solution is C = 1, and a = √ 2
The elliptic standard equation is x ^ 2 / 2 + y ^ 2 = 1 or Y ^ 2 / 2 + x ^ 2 = 1
Because the position of the long axis is not marked, there are two cases
He should not make money.
Y = - 7x & # 178; when the quadratic function x is greater than 0, the value of the function increases with the increase of the independent variable x_____ When x = 0______ When,
Y = - 7x & # 178; when the quadratic function x is greater than 0, the value of the function increases with the increase of the independent variable x_____ When x = 0______ The function has the most_____ Value, most_____ The value is_____
Y = - 7x & # 178; when the quadratic function x is greater than 0, the value of the function increases with the increase of the independent variable x__ Decrease___ When x = 0___ 0___ The function has the most___ Big__ Value, most___ Big__ The value is__ 0___
The center of the ellipse is at the origin, a focus is (0,2), passing through the point (- 3 / 2,5 / 2), and the standard equation of the ellipse is solved
C = 2, a ^ 2 = B ^ 2 + 4
So let the equation [y ^ 2 / (4 + B ^ 2)] + x ^ 2 / b ^ 2 = 1
Substituting the point coordinates, we can get B ^ 2 = 6, so a ^ 2 = 10
Y ^ 2 / 10 + x ^ 2 / 6 = 1
If the fixed-point coordinates of quadratic function y = - 2x & # 178; + BX + C are (1,2), then the value of function y decreases with the increase of independent variable x?
The vertex is (1,2)
Then the axis of symmetry is x = 1
When a = - 21, y decreases with the increase of X
The vertex coordinates of quadratic function y = - 2x & # 178; + BX + C are (1,2)
It shows that the axis of symmetry is x = 1
Open your mouth down again
Therefore, the value of function y decreases with the increase of independent variable x, and the value range of X is x > 1
The vertex coordinates are (1,2) and - 2 < 0, so the opening of quadratic function is downward,
Then the value y of the function decreases with the increase of the independent variable x in the range of X ≥ 1
Given that the hyperbola and the ellipse x ^ 2 / 9, y ^ 2 / 25 = 1 are in common focus, and the eccentricity is 2, the hyperbola equation is solved
Hyperbola C = 4, so a = 2, so B & sup2; = 12. Hyperbola is Y & sup2 / 4-x & sup2 / 12 = 1
RELATED INFORMATIONS
- 1. If (A & # 178; - 3a-2) + a of y = (A-4) x is a quadratic function, find the value of (1) a and the relation of (2) function
- 2. The value range of the independent variable in the function y = radical (x + 2) / X clearly indicates that the coefficient K of the inverse proportional function is not equal to 0. Why is the answer x > = - 2 and X ≠ 0 Value range of independent variable in function y = radical (x + 2) / X Why is the answer x > = - 2 and X ≠ 0 not x > - 2 and X ≠ 0
- 3. The image of a certain function passes through the point (- 1,2), and the value of function y decreases with the increase of independent variable x. please write a function relation that meets the above conditions___ .
- 4. If (X & # 178; - 2x + 1) of function f (x) = a is a decreasing function on (1,3), then the solution set of inequality log a (x + 1) > 0 about X is () A.{x|x>-1} B.{x|x>0} C.{x|x
- 5. F (x) is an odd function defined on R, when x > = 0 is f (x) = x ^ 2, if for any x belonging to t to t + 2, the inequality f (x + T) > = 2F (x) is constant, the value range of T is obtained
- 6. It is known that, as shown in the figure: in the plane rectangular coordinate system, O is the coordinate origin, the quadrilateral oabc is a rectangle, the coordinates of points a and C are a (10,0), C (0,4), point D is the midpoint of OA, and point P moves on the edge of BC. When △ ODP is an isosceles triangle with waist length of 5, the coordinate of point P is______ .
- 7. In the plane rectangular coordinate system, O is the coordinate origin, oabc is a rectangle, point a coordinates (10,0) point C coordinates (0,4) point D is the midpoint of OA, point P is on the BC side When the triangle ODP is an isosceles triangle with waist length 5, the coordinate of point P is?
- 8. As shown in the figure, in the right angle trapezoid oabc, the coordinates of OA ‖ CB, a and B are a (15, 0), B (10, 12) respectively. The moving points P and Q start from O and B respectively. Point P moves along OA to terminal a at a speed of 2 units per second, and point Q moves along BC to C at a speed of 1 unit per second. When point P stops moving, point Q stops moving at the same time. The line segments OB and PQ intersect at point D, and make de ‖ when passing point D Let the motion time of PQ be t (in seconds). (1) when t is the value, the quadrilateral pabq is isosceles trapezoid, please write out the reasoning process; (2) when t = 2 seconds, calculate the area of trapezoid ofBC; (3) when t is the value, △ PQF is isosceles triangle? Please write down the reasoning process
- 9. It is known that the eccentricity of ellipse C: x2 / A2 + Y2 / B2 = 1 is root sign 3 / 2, the straight line X-Y + 1 = 0 passes through the upper vertex of ellipse C, the straight line x = - 1 intersects the ellipse at two points a and B, P is different from the ellipse at any point a and B, the straight line AP BP intersects the straight line L: x = - 4 at two points Q, R, 1 respectively, to solve the equation of ellipse C, 2 to prove that the vector OQ × vector or is a fixed value
- 10. If the two focuses and the two vertices of the minor axis of the ellipse are the four vertices of a diamond with a 60 ° angle, then the eccentricity of the ellipse is 0 A. 1 / 2 b. √ 3 / 2 C. √ 3 / 3 d.1/2 or √ 3 / 2
- 11. Given y = (m ^ 2-m-2) x ^ m ^ 2-2m-1, when m is, y is the inverse proportion function of X
- 12. If the mean value + 1 of the absolute value power of K of polynomial 4xy - 1 / 5 (K-2) y is a cubic binomial, then K=
- 13. Given the function f (x) = loga (AX-1) (a > 0, a ≠ 1); (1) find the domain of definition of function f (x); (2) discuss the monotonicity of function f (x)
- 14. The image of X-2 power + 1 (a > 0, a is not equal to 1) of function y = a must pass through the point?
- 15. If the x value of quadratic function is not equal to zero, how to draw (or mark) its image?
- 16. Drawing function image with absolute value （1）y=|x-5|+|x+3| (2) Y = 2x-3 x ∈ Z | x | contained in Z （3）y=x²-2|x|-1 （4）y=｛x²+2x（x≥0） -x²-2x（x＜0） The best way to draw these four functions is to teach me. Thank you
- 17. Given the function y = 3 / (4x + 8), then the value of the independent variable x is?
- 18. Given that the image of the function f (x) = ax & # 178; + BX + C passes through the origin, for any x ∈ R, f (x-1) = f (x + 1) holds, and the equation f (x) = x has two equal real roots. Find the analytic expression of F (x) Let f (1-x) = f (1 + x) hold,
- 19. Given the function f (x) = x 2 + ax + 3, when x ∈ [- 2,2], f (x) ≥ A is constant, and the minimum value of a is obtained
- 20. If the function y = - x + M & sup2; intersects with the image of y = 4x-1 at the same point on the x-axis, the value of M is obtained 2, the first-order function y = 2 (1-k) x + 1 / 2k-1, the image does not pass through the first quadrant, find the value range of K