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It is known that the vertex coordinates of the image about the quadratic function of X are (- 1,2) and the image passes through points (1, - 3) (1) Find the analytic expression of the quadratic function image; (2) Find the intersection coordinates of the image of the quadratic function and the coordinate axis
(1) Y = ax ^ 2 + BX + C because the vertex coordinates are (- 1,2) so - B / 2A = - 1, B = 2A (4ac-b ^ 2) / 4A = 2 because B = 2A so C = 2 + a so y = ax ^ 2 + 2aX + 2 + a because the image passes through the point (1, - 3) so a = - 5 / 4 so y = - 5 / 4x ^ 2-5 / 2x + 3 / 4 (2) intersection with y axis: (0,3 / 4) intersection with X axis: root sign
When the image of quadratic function passes through points a (2,0), B (- 8, - 15 / 2) C (- 1,3), the expression is obtained
y=ax²+bx+c
be
0=4a+2b+c (1)
-15/2=64a-8b+c (2)
3=a-b+c (3)
(2)-(1)
6a-b=-3/4 (4)
(1)-(3)
a+b=-1 (5)
(4)+(5)
7a=-7/4
a=-1/4
b=-3/4
c=5/2
y=-x²/4-3x/4+5/2
It is known that the image of quadratic function intersects with X axis at two points a (- 2,0) B (1,0) and passes through point P (2,8). (1) find the expression of the quadratic function
(2) If the intersection of the parabola and the y-axis is C, find the area of △ ABC
It's not an ordinary rush!
Let y = a (x-x1) (x-x2) (x1, X2 are the two intersections of the parabola and X-axis) because the parabola passes through a (- 2,0) B (1,0), so y = a (x + 2) (x-1) because P (2,8) passes through 8 = a (2 + 2) (2-1) 8 = 4A, so a = 2, so the parabola y = 2 (x2 + X-2) y = 2x2 + 2X-4 (2) when x = 0, y = - 4, so the intersection is (0
The image of a known quadratic function is used to find the parabola through (0, - 8) (2,8) (- 1, - 22)
The image of y = ax & # 178; + BX + C passes through points (0. - 5), (1, - 8), (- 1,0)
25a-5b+c=0
a+b+c=-8
a-b+c=0
The solution is as follows
a=1
b=-4
c=-5
The analytical formula of this parabola is: y = x & # 178; - 4x-5
Typing is not easy,
RELATED INFORMATIONS
- 1. It is known that the maximum value of quadratic function is 8, and its image passes through (- 2,0) and (1,6)
- 2. Given that the symmetry axis of the image is a straight line x = - 1, and the image passes through points (1,7), (0, - 2), the analytic expression of quadratic function is obtained
- 3. Y = 1 / 2x + 1, y = 1 / 2x & # 178; - 3 / 2x + 1 find a point m on the symmetry axis of the parabola to make the value of ▏ am-mc ▏ maximum. (a is the intersection of the parabola and the Y axis, BC is the intersection of parabola and X axis, B is on the left side of C)
- 4. When the parabola y = 2x-1 is translated upward by 2 units, the analytic expression of the function is_
- 5. Parabola y = 1 / 2x & # 178; + 1 can be regarded as parabola y = 1 / 2 (x + 3) &# 178; along X axis_ Translation_ Unit
- 6. As shown in the figure a (0,4) B (2,0), C is on the positive half axis of X axis, and the parabola y = AX2 + BX + C passes through ABC three points
- 7. The axis of symmetry of the parabola y = ax ^ 2 + BX + C is x = 2, passing through (0,4) and (5,9) Write down the whole process
- 8. It is known that the image vertex coordinates of the quadratic function are a (- 2,0) and pass B (- 1,2) (find the analytic formula of the quadratic function, let the image intersect with the Y axis at point C, and find the area of △ ABC It is known that the image vertex coordinates of the quadratic function are a (- 2,0) and pass B (- 1,2) (1) to find the analytic formula of the quadratic function. (2) let the image intersect with the Y axis at point C, and find the area of △ ABC
- 9. The parabola y = AX2 + BX + C passes through point a (3,0). B (2, - 3) with x = 1 as the symmetry axis Is there a point P on the symmetry axis X = 1 so that PA = Pb in the triangle PAB? If so, find out the coordinates of point P. if not, explain the reason
- 10. Given that the vertex coordinates of a parabola are (- 2,1) and pass through points (1, - 2), the analytic formula of the parabola is obtained
- 11. If the coordinates of three vertices of triangle ABC are a (- 3, - 1), B (2, - 1), C (1,3), then the area of triangle ABC is
- 12. Given that the image of quadratic function y = 2x square + BX + C passes through point a (2,3) and intersects with line y = 3x-2 at point B (1, m), the relation of quadratic function is obtained
- 13. For quadratic function y = - 2x ^ 2, find the value range of y when - 1 ≤ x ≤ 2
- 14. How to solve 9x-12 * 2 = 3x
- 15. （3x-4)^2=9X-12
- 16. Draw the image of function y = 3x-15, observe the image and answer: when x = what value, y = 0? Y > 0? Y < 0? Must be a graph,
- 17. Given the function y = Y1 + Y2, and Y1 = 2x + m, y2 = x / (m-1) + 3, the ordinate of the image focus of the two functions Y1 and Y2 is 4
- 18. As shown in the figure, the functional relations of lines OC and BC are Y1 = x and y2 = - 2x + 6, respectively. The moving point P (x, 0) moves on ob (0 < x < 3), and the line m passing through point P is perpendicular to the X axis. (1) find the coordinates of point C, and answer when x takes what value, Y1 > Y2? (2) Let the area of the left part of the line m in △ cob be s, and the functional relationship between S and X is obtained. (3) when x is what, the line m bisects the area of △ cob?
- 19. When m is the value, the focus of the line y = 5x + m on the line y = 3x-4 is on the x-axis?
- 20. It is known that: as shown in the figure, in the rectangular coordinate system xoy, one side of RT △ OCD OC is on the X axis. ∠ C = 90 °, point D is in the first quadrant, OC = 3, DC = 4, and the image of the inverse scale function passes through the midpoint a of OD. (1) find the analytical formula of the inverse scale function; (2) if the image of the inverse scale function intersects with the other side of RT △ OCD DC at point B, find the analytical formula of the straight line of a and B