(1) Try to use an algebraic expression containing m to represent the symmetry axis of this quadratic function image.
(2) If tan∠PAB=2, find the coordinates of the vertex P of this quadratic function image.
(3) Under the condition of sub-item (2), let the image of this function intersect with the Y axis at point C, and try to compare the sizes of ∠APC and ∠ABC to prove your guess.
(1) Analysis: The image of ∵ quadratic function y =- 1/2x 2+bx+c passes through point A (-6,0) and point B (m,0 0) (m > 0).
Substitute a and b coordinates into-18-6b =-1/2m 2+BM = > b = (m 2-36)/(2m+12).
Symmetry axis of function image: x = b = (m 2-36)/(2m+12) = (m-6)/2.
(2) Analysis: Let the vertex of the function image be P((m-6)/2, y0), and tan∠PAB=2.
∴y0={(m-6)/2+6}tan∠PAB =m+6
∴P((m-6)/2,m+6)
∴y=- 1/2(x+6)(x-m)=- 1/2x^2+(m-6)/2x+3m=- 1/2[x-(m-6)/2]^2+3m+(m-6)^2/8
Its maximum value is 3m+(m-6) 2/8.
Let 3m+(m-6) 2/8 = m+6 = > m1= 2, m2=-6 (s).
∴P(-2,8)
(3) Analysis: ∵ Function y =- 1/2x 2-2x+6
∴A(-3,0),B(2,0),C(0,6),P(-2,8)
At ⊿APC
AP^2=80,PC^2=8,AC^2=72
∴ap^2=pc^2+ac^2==>； PC⊥AC
tan∠APC=AC/PC=3
At the ⊿OBC
OC=6，OB=2
∴tan∠OBC=OC/OB=3
∴∠APC=∠ABC