由于函数图像经过A(4,0)B(-4,-4)两点,这两点应符合函数特性,则根据题义得方程组,
y = - x^2/4 + 4b + c = 0 ①
y = - x^2/4 - 4b + c = -4②
解之后可得b = 1/2c = 2
函数为y = - x^2/4 + x/2 + 2 = 0图像如插图
tan ∠CAO= |CO|/|AO| = 2/4 = 1/2
tan ∠BAO= 4/8 = 1/2
tan ∠CAO = tan ∠BAO
所以 ∠CAO = ∠BAO
由于函数图像经过A(4,0)B(-4,-4)两点,这两点应符合函数特性,则根据题义得方程组,
y = - x^2/4 + 4b + c = 0 ①
y = - x^2/4 - 4b + c = -4②
解之后可得b = 1/2c = 2
函数为y = - x^2/4 + x/2 + 2 = 0图像如插图
tan ∠CAO= |CO|/|AO| = 2/4 = 1/2
tan ∠BAO= 4/8 = 1/2
tan ∠CAO = tan ∠BAO
所以 ∠CAO = ∠BAO