http://x85.xanga.com/29ac306553232152578852/b113890195.jpg
最尾果幅圖
The figure shows the graph of y = x 2次+ px +q which cuts the x-axis at A(1, 0)
and B(6,0).FIND
(a)the values of p and q.
(b)the optimum value of x2次 +px+q.
中4數學問題 有20分嫁~
2007-10-19 2:51 am
回答 (2)
2007-10-19 3:18 am
✔ 最佳答案
Since x=1 and x=6 is the root of y,Hence, (x-1)(x-6)=x^2+px+q
...........x^2-7x+6=x^2+px+q
Therefore, p=-7, q=6
y=x^2-7x+6
..=[x^2-7x+(7/2)^2]-(7/2)^2+6
..=(x-7/2)^2-25/4
Hence, optimal value of y is -25/4, occur at x=7/2
2007-10-19 3:22 am
(a)
y = x^2 + px + q
代入 (1,0)
0 = 1^2 + p + q
p = -1 - q ------------- (1)
代入 (6,0)
0 = 6^2 + 6p + q
0 = 36 + 6p + q ---- (2)
(1) 代入 (2)
0 = 36 + 6(-1-q) + q
0 = 36 - 6 - 6q + q
0 = 30 - 5q
q = 6 #
q 代入 (1)
p = - 1 - 6
p = -7 #
所以方程:
y = x^2 - 7x + 6
(b)
配方法:
y = x^2 - 7x + 6
y = x^2 - 7x + (7/2)^2 - (7/2)^2 + 6
y = x^2 (x + 7/2)^2 - 25/4
所以 x 極大值 = 7/2
y = x^2 + px + q
代入 (1,0)
0 = 1^2 + p + q
p = -1 - q ------------- (1)
代入 (6,0)
0 = 6^2 + 6p + q
0 = 36 + 6p + q ---- (2)
(1) 代入 (2)
0 = 36 + 6(-1-q) + q
0 = 36 - 6 - 6q + q
0 = 30 - 5q
q = 6 #
q 代入 (1)
p = - 1 - 6
p = -7 #
所以方程:
y = x^2 - 7x + 6
(b)
配方法:
y = x^2 - 7x + 6
y = x^2 - 7x + (7/2)^2 - (7/2)^2 + 6
y = x^2 (x + 7/2)^2 - 25/4
所以 x 極大值 = 7/2
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https://hk.answers.yahoo.com/question/index?qid=20071018000051KK02539