Consider the function f(x)=x^2-ax-(a+1), where a=/= -2. The equation f(x)=0 has
two distinct real roots. If the y-intercept of the quadratic graph y=f(x) is 5, find
the x-intercepts of the graph.
quadratic equations in one unk
2012-09-08 4:38 am
回答 (1)
2012-09-08 4:54 am
✔ 最佳答案
y = x² - ax - (a + 1)When x = 0, y = 5 :
5 = 0 - 0 - (a + 1)
a + 1 = -5
a = -6
Hence, f(x) = x² + 6x + 5
When y = 0 :
0 = x² + 6x + 5
(x + 1)(x + 5) = 0
x = -1 or x = -5
The y -intercepts : -1 and -5
參考: 賣女孩的火柴
收錄日期: 2021-04-13 18:57:41
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