It is given that the graph passes through the origin and its vertex is (2,-1).Find f(x)
個答案係1/4x^2-x
更新1:
個答案係1/4x^2-x !!!!!!! PLZ HELP ME!!!!
中四數學(Quadratic Function)
2010-01-02 11:15 pm
The graph of a Quadratic Function y=f(x).
It is given that the graph passes through the origin and its vertex is (2,-1).Find f(x)
個答案係1/4x^2-x
It is given that the graph passes through the origin and its vertex is (2,-1).Find f(x)
個答案係1/4x^2-x
回答 (3)
2010-01-02 11:31 pm
✔ 最佳答案
let the equation bef(x) = ax^2 + bx + c
as it pass through the origin
f(0) = 0
0 = a(0)^2 + b(0) + c
c = 0
vertex of quadratic equation = -1
-b/(2a) = -1
b = 2a
f(2) = -1
-1 = a(2)^2 + b(2)
-1 = 4a + 2b
-1 = 2b + 2b
b = -1/4
a = -1/8
f(x) = (-1/8) (x^2 + 2x)
2010-01-03 1:59 am
但係個答案係1/4x^2-x
2010-01-03 1:40 am
It is easier to use the form y = a(x + h)^2 + k.
Since vertex is ( 2, -1), so 2 + h = 0, h = - 2 and k = -1.
That is y = a( x - 2)^ - 1.
For x = 0, y = 0,
so 0 = a( 0 - 2)^2 - 1
0 = 4a - 1
a = 1/4
so the function is y = 1/4(x - 2)^1 - 1 = 1/4(x^2 - 4x + 4) - 1
= x^2/4 - x + 1 - 1 = x^2/4 - x.
Since vertex is ( 2, -1), so 2 + h = 0, h = - 2 and k = -1.
That is y = a( x - 2)^ - 1.
For x = 0, y = 0,
so 0 = a( 0 - 2)^2 - 1
0 = 4a - 1
a = 1/4
so the function is y = 1/4(x - 2)^1 - 1 = 1/4(x^2 - 4x + 4) - 1
= x^2/4 - x + 1 - 1 = x^2/4 - x.
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