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The seventh mathematical finale equation
Analysis: (1) Substitute x=0 into the analytical formula of parabola;

(2) When b=0 and the straight line is y=x, the coordinates of b and c can be obtained by solving the equations composed of y=x and y=x2+x-4, and then the area can be obtained by using the triangle area formula; (3) When b >-4, the areas of △ABE and △ACE are equal, because the coordinates of the intersection point can be obtained by solving the equations composed of straight lines and parabolas, and the BF⊥y axis, CG⊥y axis and vertical foot are F and G respectively. According to the coordinates of points, two triangles with the same base can be obtained, and the answer can be obtained;

(4) There is such a B. According to congruent triangles's judgment, it is proved that △ BEF △ CEG is deduced to BE=CE. According to the properties of right triangle, when OE=CE, △OBC is a right triangle, and the value of b can be obtained by substitution. Solution: (1) solution: substitute x=0 into the analytical formula of parabola to get: y.

The coordinates of point A are (0, -4),

Answer: The coordinate of point A is (0, -4).

(2) Solution: When b=0, the straight line is y=x,

By {y=xy=x2+x-4,

The solution is {x 1=2y 1=2, {x2=-2y2=-2,

∴ The coordinates of B and C are B(-2, -2) and C(2, 2) respectively.

S△ABE= 12×4×2=4,S△ACE= 12×4×2=4,

Answer: the area of △ABE is 4, and the area of △ACE is 4.

(3) solution: when b >-4, S△ABE=S△ACE,

The reason is {y=x+by=x2+x-4,

The solution is {x 1=b+4y 1=b+4+b, {x2=-b+4y2=-b+4+b,

∴ The coordinates of b and c are:

B(- b+4,- b+4+b),C( b+4,b+4+b),

Let BF⊥y axis, CG⊥y axis and vertical foot be f and g respectively.

Then BF=CG=b+4,

And △ABE and △ACE are two triangles with the same base.

∴S△ABE=S△ACE.

A: when B >-4, the area relationship between △ABE and △ACE is equal.

(4) solution: there is such a b,

BF = CG,∠BEF=∠CEG,∠BFE=∠CGE=90,

∴△BEF≌△CEG,

∴BE=CE,

That is, e is the midpoint of BC,

So when OE=CE, △OBC is a right triangle.

∫GE = b+4+b-b = b+4 = GC,

∴ CE=2? B+4 and OE=|b|,

∴ 2? b+4=|b|,

The solution is b 1=4. b2=-2,

When b=4 or -2, △OBC is a right triangle.

A: There is a right triangle with B△BOC as the hypotenuse, and the value of B is 4 or -2.