f.4 maths : functions and graphs (I)

1. Let y = 3x + 2, so x = (y-2)/3.
Therefore, x^2 - x - 1 = (y-2)^2/9 - (y-2)/3 - 1.
Therefore, f(y) = (y-2)^2/9 - (y-2)/3 - 1. Since y is a dummy variable, we can re-write to
f(x) = (x-2)^2/9 - (x-2)/3 - 1.
So f(0) = 4/9 - (-2/3) - 1 = 4/9 + 2/3 - 1 = (4 + 6 - 9)/9 = 1/9.
2.
For y = ax^2 + bx + c.
y - intercept = c.
x - intercepts are the roots of the equation ax^2 + bx + c = 0.
Vertex is (-b/2a, -delta/4a), where delta is b^2 - 4ac.
With these points located, you can draw the graph.
3.
Let the inequality be f(x) > k. where y = f(x) is the graph and k is a constant.
Step 1 : Draw the horizontal line y = k.
Step2: Mark the intersecting points of y = k and y = f(x).
So the portions of f(x) above y = k will be f(x) > k. The portions of f(x) below y= k will be f(x) < k.
Once these portions are determined, the range of x for these portions can be found from the x - axis.

2008-10-25 08:27:18 補充：
The last line: ....range of x can be found from the x -axis according to the locations of the intersecting points.