幾條maths question

Q1.
y = x^2 + bx + 14 = (x + b/2)^2 - b^2/4 + 14.
y minimum = - b^2/4 + 14 = 10
4 = b^2/4
16= b^2
b = +/- 4 .
Q2.
When a polynomial is divided by ( x +2), the remainder should not contain term with x because division can go on, so some information may be missing, please check.
Q3.
Let the line be y = mx + c
when x = 0, y = -1 so - 1 = c
when y = 0, x = 4, so 0 = 4m - 1
m = 1/4
so the line is y = x/4 - 1
or 4y = x - 4.
Q4.
Slope of the required line = slope of given line = 1/8 since they are parallel.
So equation of the required line is
y - (-2) = 1/8( x - 3)
y + 2 = (x - 3)/8
8y + 16 = x - 3
x - 8y - 19 = 0.
Q5.
x^2 + y^2 - 4x - 8y + 11 = 0
(x - 2)^2 - 4 + (y - 4)^2 - 16 + 11 = 0
(x - 2)^2 + (y - 4)^2 = 9
so center is ( 2, 4) and radius = 3.
Distance between (2, 4) and (0,0) = sqrt [ 2^2 + 4^2] = sqrt ( 4 + 16)
= sqrt 20 > 3, therefore (0,0) is outside the circle.
so 1,2 and 3 are true.



2009-07-09 07:29:55 補充：
For Q2, you can say that f(x) = (x + 2)(4x^2 - 1) + (3 - 2x) = 4x^3 + 8x^2 - 3x + 1
but (3 - 2x) is not really the remainder because it can still be divided by ( x + 2).