It is given that the parabola P1 : y= -(x-h)^2 +k passes through A(-4,-1). B(0,-1) is the image of A with respect to the axis of symmetry. The parabola P2 : y= mx^2 -3x -5 passes through A.
a) Find the values of h,k and m.
b) Determine which of the two parabolas has a wider opening.
Quadratic
2008-02-16 9:02 pm
回答 (1)
2008-02-16 9:26 pm
✔ 最佳答案
(a) In the equation of P1, sub x = 0 and -4 will both give y = -1 respectively.So,
-1 = -(-h)^2 + k
-h^2 + k = -1 ... (1)
-1 = -(4 - h)^2 + k
-h^2 + 8h - 16 + k = -1
-h^2 + 8h + k = 15 ... (2)
Solving (1) and (2), we have:
h = 2 and k = 3
Sub x = -4 into the equation of P2 will give y = -1:
-1 = 16m + 12 - 5
m = -1/2
(b) Coefficient of x^2 of P1 = -1
Coefficient of x^2 of P2 = -1/2
Therefore P2 has a wider opening as its coefficient of x^2 has a smaller absolute value.
參考: Myself
收錄日期: 2021-04-14 19:43:17
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