[中四]函數

http://xs219.xs.to/xs219/07386/6745754645.jpg
1)
g(x) = 6 / (x-2)
Therefore g(3) = 6/(3-2) = 6/1 = 6 (Put x=3 into g(x))
Similarly, g(-4) = 6/(-4-2) = 6/(-6) = -1
Thus, g(3) - g(-4) = 6-(-1) = 6+1 = 7

2)
f(x) = x^2 - 2x - 5
f(k+1) = (k+1)^2 - 2(k+1) - 5
= k^2 + 2k + 1 - 2k - 2 - 5
= k^2 - 6
Since f(k+1) = k
k^2 - 6 = k
k^2 - k - 6 = 0
(k-3)(k+2) = 0
Therefore, k=3 or k=-2
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http://xs219.xs.to/xs219/07386/44444444444444.jpg
a)
y = 3(x-2)^2 + 7
According to the equation, y = a(x-h)^2 + k, x = h will be the equation of 對稱軸
Therefore, x = 2
b)
If x = h, y will be the minimum
Therefore here x = 2
The minimum value of y = 3(2-2)^2 + 7
y = 7
c)
The coordinates of the vertex must be the lowest point of the curve
As the figure shows, we know that that point passes through the 對稱軸
Therefore we can put x=2 into the equation y = 3(x-2)^2 + 7
Thus
y = 3(2-2)^2 + 7
y = 7 (As same as the minimun value of y)
Therefore the vertex is (2,7)

g(x)=6/(x-2)

g(3)-g(-4)
=6/1-6/(-6)
=6-1
=5

y=3(x-2)^2+7............(#)

對程軸=
x=2

(x=2)代入(#),
∴y的極小值=3

頂點=(2,3)

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