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Test and analysis: (1) First, the analytical formula of straight line AB is y=x+3, and the coordinates of its intersection point A with X axis and its intersection point B with Y axis are obtained, then the coordinates of A and B are substituted into y=-x2+bx+c, and the analytical formula of parabola can be obtained by using the undetermined coefficient method;
(2) Let the coordinates of point F in the third quadrant be (m, -m2-2m+3), find the coordinates of parabola symmetry axis and vertex D by matching method, and then let the parabola symmetry axis and X axis intersect at point G to connect FG. According to S△AEF=S△AEG+S△AFG-S△EFG=3, the equation about m is listed and solved.
(3) The coordinate of point P is (-1, n). Starting from the coordinates of point B and point C, we can get BC2= 10 by pythagorean theorem, and then discuss it in three situations: ① ∠ PBC = 90, firstly we can get PB2+BC2=PC2 by pythagorean theorem, and then we can list the equation ② ∠ BPC = 90 about n, and we can get the corresponding t value by ①; ③∠BCP = 90°, and the corresponding t value can be obtained by ①.
Test analysis: (1)∵y=x+3 intersects with X axis at point A, and intersects with Y axis at point B,
When y=0, x=-3, that is, the coordinate of point A is (-3,0).
When x=0 and y=3, the coordinate of point B is (0,3).
Substituting a (-3,0) and b (0 0,3) into y=-x2+bx+c, we get.
, solutions,
The analytical formula of parabola is y =-x2-2x+3;
(2) As shown in figure 1,
Let the coordinate of point F in the third quadrant be (m, -m2-2m+3), then m < 0, -m2-2m+3 < 0.
∫y =-x2-2x+3 =-(x+ 1)2+4，
∴ Symmetry axis is a straight line x=- 1, and the coordinate of vertex d is (-1, 4).
Let the symmetry axis of parabola intersect with the X axis at point G and connect FG, then G (- 1, 0) and Ag = 2.
The analytical formula of ∫ straight line AB is y=x+3,
When x=- 1 and y=- 1+3=2,
∴ The coordinate of point E is (-1, 2).
∫S△AEF = S△AEG+S△AFG-S△EFG =×2×2+×2×(m2+2m-3)-×2×(- 1-m)= m2+3m，
∴ When the area of a triangle with vertices A, E and F is 3, m2+3m=3,
Solution: (give up),
At that time, -m2-2m+3=-m2-3m+m+3=-3+m+3=m=, ∴ the coordinate of point F is (,);
(3) The coordinate of point P is (-1, n).
∫B(0，3)，C( 1，0)，
∴BC2= 12+32= 10.
There are three situations: ① As shown in Figure 2, if ∠ PBC = 90, PB2+BC2=PC2.
Namely, (0+1) 2+(n-3) 2+10 = (1+1) 2+(n-0) 2,
Simplify the arrangement to get 6n= 16, and get n=,
∴ The coordinate of point P is (-1,),
∫ The coordinate of vertex d is (-1, 4),
∴PD=4-=，
The speed of point P is 1 unit length per second.
∴t 1=；
② As shown in Figure 3, if ∠ BPC = 90, PB2+PC2=BC2.
That is, (0+1) 2+(n-3) 2+(1+1) 2+(n-0) 2 =10,
N2-3n+2=0 after simplification, and n=2 or 1 after solution.
∴ The coordinates of point P are (-1, 2) or (-1, 1).
∫ The coordinate of vertex d is (-1, 4),
∴PD=4-2=2 or PD=4- 1=3,
The speed of point P is 1 unit length per second.
∴t2=2,t3=3；
③ As shown in Figure 4, if ∠ BCP = 90, then BC2+PC2=PB2.
Namely10+(1+1) 2+(n-0) 2 = (0+1) 2+(n-3) 2,
Simplify the arrangement to get 6n=-4, and solve it to get n=-.
∴ The coordinate of point P is (-1,-),
∫ The coordinate of vertex d is (-1, 4),
∴PD=4+=，
The speed of point P is 1 unit length per second.
∴t4=；
To sum up, when t is seconds or 2 seconds or 3 seconds or seconds, the triangle with P, B and C as vertices is a right triangle.