The abscissa of vertex c is 4, which passes through point (0,).
∴y=a(x-4)2+k……①
In addition, the symmetry axis is a straight line x=4, and the line segment length of the image cut on the X axis is 6.
∴A( 1,0),B(7,0)
∴0=9a+k……②
A= and k= are obtained from ① ② solution.
The analytic formula of the quadratic function is: y = (x-4) 2-
(2) Point A and Point B are symmetrical about the straight line x=4.
∴PA=PB
∴PA+PD=PB+PD≥DB
When the point P is on the line segment DB, PA+PD takes the minimum value.
The intersection of db and symmetry axis is point p.
Let the straight line x=4 and the x axis intersect at point m.
Pm od, ∴∠BPM=∠BDO and ∠PBM=∠DBO.
∴△BPM∽△BDO
∴ ∴
The coordinate of point p is (4,)
(3) From (1), we can know that point C(4,),
Am = 3, ∴△AMC, cot∠ACM= in Rt,
∴∠acm=60o,∵ac=bc,∴∠acb= 120o
(1) when the point q is above the x axis, q is qn ⊥ and the x axis is n.
If AB=BQ, it is given by △ABC∽△ABQ.
BQ=6, ∠ABQ= 120o, then ∠QBN=60o.
∴QN=3,BN=3,ON= 10,
At this time Q( 10,),
If AB=AQ, Q(-2,) is known from symmetry.
② When the Q point is below the X axis, △QAB is △ACB.
At this time, the coordinate of point q is (4,),
After investigation, the points (10,) and (-2,) are all on the parabola.
To sum up, there is such a point Q, that is, △QAB∽△ABC.
The coordinates of point Q are (10,) or (-2,) or (4).