(b)When x=-4
y=a(-4)^2+2(a+1)(-4)+a+3
=16a-8a-8+a+3
=9a-5
>9x1-5,i.e >4
P(-4,4)does not lie on the graph .
(why?a>1,4>1,唔係咩?)
更新1:
如果P(-4,5)係唔係條線上? 係,因為5>4,係唔係咁?
F.4 Maths
2011-12-30 12:14 am
(a) The range of possible values of a is a>1
(b)When x=-4
y=a(-4)^2+2(a+1)(-4)+a+3
=16a-8a-8+a+3
=9a-5
>9x1-5,i.e >4
P(-4,4)does not lie on the graph .
(why?a>1,4>1,唔係咩?)
(b)When x=-4
y=a(-4)^2+2(a+1)(-4)+a+3
=16a-8a-8+a+3
=9a-5
>9x1-5,i.e >4
P(-4,4)does not lie on the graph .
(why?a>1,4>1,唔係咩?)
回答 (2)
2011-12-30 12:48 am
✔ 最佳答案
不太明你的意思。代 x = 4 時 y = 9a - 5
因為a > 1﹐因此y > 9 * 1 - 5 = 4
(這個4是對y值說的﹐不是對a說的)
因此P(-4,4)一定不在y = f(x) 上
希望幫到你
2011-12-30 10:01:43 補充：
不是。因為都不知道y的真確值。例如 y 可以是5,6,7,8,9,10...
2011-12-30 2:48 am
y = f(x) = ax^2 + 2(a+1)x + (a+3)
y > 0 for all values of x => f(x) doesn't touch or cut x-axis
=> discriminant < 0
[2(a+1)]^2 - 4a(a+3) < 0
4(a^2 + 2a + 1) - 4a^2 - 12a < 0
-4a + 4 < 0
4 < 4a
1 < a
when x = -4
y = 16a - 8a - 8 + a + 3
y = 9a - 5
if y = 4, a = 1 but it contradicts to the question as a is greater than 1
so, (-4, 4) does not lie on the graph
y > 0 for all values of x => f(x) doesn't touch or cut x-axis
=> discriminant < 0
[2(a+1)]^2 - 4a(a+3) < 0
4(a^2 + 2a + 1) - 4a^2 - 12a < 0
-4a + 4 < 0
4 < 4a
1 < a
when x = -4
y = 16a - 8a - 8 + a + 3
y = 9a - 5
if y = 4, a = 1 but it contradicts to the question as a is greater than 1
so, (-4, 4) does not lie on the graph
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