26.( 1)=——(x— 10)2+900；

(2) In the first 20 days, when x= 10, R 1 is the largest, which is 900 yuan.

On the last day of 10, when x=2 1, R2 is the largest, and it is 950 yuan.

Therefore, in the 30-day trial sales, the daily sales profit of 2 1 day is the largest, and it is 95 yuan.

27.( 1) OG ⊥ CD, which proves that if OC and OD are connected, OC=OD, because G is the midpoint of CD, and OG ⊥ CD;

(2)∫ arc CD= arc CD ∴∠CAE=∠CBF, and ∫≈ace =∠BCF = 90? ，AC=BC，

∴△cae≌△cbf ∴ae=bf；

(3)∵∠CAD =∠ bud ∴ arc CD= arc DB ∴CD=DB

∫∠ACB = 90？ ∴AB is diameter∴∠ ADB = 90? ,

∵OG⊥CD ∴∠CGO=∠BDE=90？

∫△ABC is an isosceles right triangle ∴∠OCB=∠CBO=45?

∫ Arc decibel = Arc decibel ∴∠DAB=∠DCB

∴∠dcb+∠ocb=∠dab+∠cbo,∴∠ocg=∠dea

∴△ocg∽△dbe ∴og∶db=cg∶de ∴db×cg=og×de=

∫CG = CD = db ∴db2=

∵∠ACE=∠ADB=90？ ，∠CAE =∠dab ∴△ace∽△adb ∴ac∶ad=ae∶ab

∴AC×AB=AE×AD, let the radius of⊙ O be R, then AC= R and AB=2R.

∴AE×AD= R2，

Arc CD= arc DB ∴∠DBE=∠DAB, ∫∠bde =∠ADB = 90?

∴△dbe∽△dab ∴db∶da=de∶db ∴db2=da×de

∴AE×AD+DA×DE= R2+DB2, that is, AD2= R2+DB2.

∫ad2 = ab2—DB2 ∴4r2—db2=r2+db2,∴4r2—R2 = 2d B2 =

∴ R2 = 6 ∴⊙ The area of O O is 6π.

28.( 1) from the title, C(0, -3), from cos∠BCO=, get B (1, 0).

Substituting the coordinates of b and c, we get -3 = A+C, 0=4a+c, and we get a= 1, and c =-4.

So the parabola is y = (x+ 1) 2-4.

(2) From (1), we know m (- 1, -4). If we substitute it into the straight line MC, we get -4 =-k-3, so k= 1.

The straight line MC is y = x-3, so n (3 3,0)

WhEN n is the vertex of a right angle, the perpendicular en passing through n and MC intersects the parabola at P 1, P2, en: y =-x+3.

Simultaneous parabolic expressions can give P 1 (,), P2 (,).

When c is the vertex of a right angle, P3 (-3,0) if the vertical line FC passing through c as MC intersects the parabola at another point P3.

To sum up, there is a point P on the parabola that is different from point C, and its coordinate is (,).

Or (,) or (-3,0)

Q (-3) Let the translated parabola be y=(x+ 1)2+k, q (-3, -6).

By (x+ 1) 2+k = x-3, that is, x2+x+k+4=0, △ = 12-4 (k+4) ≥ 0, and k ≤

At most 4- 15/4 = 1/4 unit length.

When crossing point n, 0= 16+k, k =- 16, when crossing point q, -6 = 4+k, k =- 10,

Therefore, the maximum downward translation is 16-4 = 12 unit lengths.