As shown in the figure, the image of the quadratic function y = ax2-4x + C passes through the coordinate origin and intersects with the X axis at point a (- 4,0) (1) Find the analytic formula of quadratic function; (2) If there is a point P on a parabola, s △ AOP = 8, please write out the coordinates of point P directly

As shown in the figure, the image of the quadratic function y = ax2-4x + C passes through the coordinate origin and intersects with the X axis at point a (- 4,0) (1) Find the analytic formula of quadratic function; (2) If there is a point P on a parabola, s △ AOP = 8, please write out the coordinates of point P directly

(1) According to the known conditions, C = 0A × (- 4) 2-4 × (- 4) + C = 0, and a = - 1C = 0. Therefore, the analytic formula of the quadratic function is y = - x2-4x; (2) ∵ the coordinates of point a are (- 4, 0),

As shown in the figure, we know that the image of the quadratic function y = ax2-4x + C passes through point a and point B. find the analytic formula of the quadratic function; Write the symmetry axis and vertex coordinates of the parabola!!!! Thank you

(1) By substituting x = - 1, y = - 1; X = 3, y = - 9 into y = ax2-4x + C respectively, the expression of the quadratic function is y = x2-4x-6

As shown in the figure, the image of the quadratic function y = ax2-4x + C passes through the coordinate origin and intersects with the X axis at point a (- 4,0) (1) Find the analytic formula of quadratic function; (2) If there is a point P on a parabola, s △ AOP = 8, please write out the coordinates of point P directly

(1) From the known conditions
C=0
a×(-4)2-4×(-4)+c=0 ，
The solution
a=-1
c=0 ，
Therefore, the analytic expression of the quadratic function is y = - x2-4x;
(2) ∵ the coordinates of point a are (- 4, 0),
∴AO=4，
Let the distance from point P to X axis be H,
Then s △ AOP = 1
2×4h=8，
The solution is h = 4,
① When the point P is above the X axis, - x2-4x = 4,
X = - 2,
Therefore, the coordinates of point P are (- 2,4),
② When the point P is below the X axis, - x2-4x = - 4,
The solution is X1 = - 2 + 2
2，x2=-2-2
2，
So, the coordinates of point P are (- 2 + 2)
2, - 4) or (- 2-2)
2，-4），
To sum up, the coordinates of point P are: (- 2,4), (- 2 + 2)
2，-4）、（-2-2
2，-4）．

As shown in the figure, we know that the vertex of the image of the quadratic function y = x2-2x-1 is a, and the image of the quadratic function y = AX2 + BX + C intersects the origin O and another point C with the X axis. Its vertex B is on the symmetry axis of the image of the function y = x2-2x-1 (1) Find the coordinates of point a and point C; (2) When point B and point a are symmetric about the X axis, the analytic formula of the function y = AX2 + BX + C is obtained, and whether the quadrilateral aobc can pass through a circular hole with a diameter of 1.8 is determined

(1) ∵ y = x2-2x-1, ? y = (x-1) 2-2, ᙽ a (1, - 2), ? y = AX2 + BX + C is on the symmetry axis of the image of the function y = x2-2x-1, as shown in the figure: ? of = 1 according to the symmetry of parabola, FC = 1,  co = 2, ? C (2,0); (2) ∵ point B and point a are symmetric about the x-axis ? B (1, 2) ? B (1,2)

As shown in the figure, the radius OA of the center of circle O is known to be OA = 5, the point C is the point on the chord AB, CO is perpendicular to OA and OC = BC, and find the length of ab Miscellaneous work

∵OA=OB
∴∠OAB=∠OBA
∵OC=BC
∴∠COB=∠OBA=1/2=∠OCA
∵OC⊥OA
∴∠OAB=∠OBA=∠COB=30°
∴OA=√3OC,AC=2OC
∴OC=5/√3
∴AB=3OC=5√3

As shown in the figure, OA is the radius of ⊙ o

Proof: connect od and be,
∵ OA and OE are the diameters of ⊙ C and ⊙ o respectively,
∴∠ADO=∠ABE=90°，
∴OD∥BE，
∵ o is the midpoint of AE,
﹤ D is the midpoint of ab

As shown in the figure, OA is the radius of ⊙ o

Proof: connect od and be,
∵ OA and OE are the diameters of ⊙ C and ⊙ o respectively,
∴∠ADO=∠ABE=90°，
∴OD∥BE，
∵ o is the midpoint of AE,
﹤ D is the midpoint of ab

As shown in the figure, OA is the radius of ⊙ o

Proof: connect od and be,
∵ OA and OE are the diameters of ⊙ C and ⊙ o respectively,
∴∠ADO=∠ABE=90°，
∴OD∥BE，
∵ o is the midpoint of AE,
﹤ D is the midpoint of ab

As shown in the figure, in ⊙ o, AB is the chord, C is the midpoint of arc AB, OC intersects AB with D, ab = 6cm, CD = 1cm, calculate the radius OA of ⊙ o

∵ C is the midpoint of arc ab,
∵AB⊥OC，
∵AB=6cm，
∴AD=1
2AB=3cm，
Let OA = R, then od = r-cd = R-1,
In RT △ AOD,
∵ oa2 = ad2 + OD2, that is, R2 = 32 + (R-1) 2,
The solution is r = 5

As shown in the figure, the radius OA of the center of circle O is known to be OA = 5, the point C is the point on the chord AB, CO is perpendicular to OA and OC = BC, and find the length of ab

Make auxiliary line, extend Bo to D, connect ad, use triangle AOC to be similar to triangle bad, list the relationship between AB and OA equal to BD to AC,