MATHS.(好急!!!!!求下你地幫手!)

Question 1a
y = x(1 - x) - k / 4
y = -x^2 + x - k / 4
Because the graph cut x-axis at one point,
-x^2 + x - k / 4 = 0 has one root,
b^2 - 4ac = 0
(1)^2 - 4(-1)(-k / 4) = 0
1 - k = 0
k = 1

Question 1b
the formula is y = -x^2 + x - 1 / 4
for the x-intercept, sub. y = 0,
-x^2 + x - 1 / 4 = 0
4x^2 - 4x + 1 = 0
(2x + 1)^2 = 0
x = -1 / 2
so the coordinates of Q is (-1/2, 0)

Question 2a
y = (x - 4p)(x - 2) + p
y = x^2 - 4px - 2x + 8p + p
y = x^2 + (- 4p - 2)x + 9p
because the graph has one x-intercept,
x^2 + (- 4p - 2)x + 9p = 0 has one root,
b^2 - 4ac =0
(- 4p - 2)^2 - 4(1)(9p) = 0
16p^2 +16p + 4 - 36p = 0
16p^2 - 20p + 4 = 0
4p^2 - 5p + 1 = 0
(4p - 1)(p - 1) =0
p = 1/4 or 1

Question 2b
For p = 1/4,
y = x^2 + [- 4(1/4) - 2]x + 9(1/4)
y = x^2 - 3x + 9/4
for the x-intercept, sub. y = 0,
x^2 - 3x + 9/4 = 0
4x^2 - 12x + 9 = 0
(2x - 3)^2 = 0
x = 3/2
so for p = 1/4, the x-intercept is (3/2, 0)
For p = 1,
y = x^2 + [- 4(1) - 2]x + 9(1)
y = x^2 - 6x + 9
for the x-intercept, sub. y = 0,
x^2 - 6x + 9 = 0
(x - 3) = 0
x = 3
so for p = 1, the x-intercept is (3, 0)

2009-10-04 03:41:31 補充：
Question 3
y = 9(x - 1)^2 - m - 6
y = 9(x^2 - 2x + 1) - m - 6
y = 9x^2 - 18x + 9 - m - 6
y = 9x^2 - 18x - m + 3
because the graph intersects the x-axis,
b^2 - 4ac >= 0
(- 18)^2 - 4(9)(- m + 3) >= 0
324 + 36m - 108 >= 0
216 + 36m >= 0
36m >= - 216
m >= - 6
so that the smallest value of m is 6

2009-10-04 03:41:45 補充：
Question 4a
y = - 2x^2 + 3x - k - 1
because the graph cut the x-axis at 2 points,
b^2 - 4ac > 0
(3)^2 - 4(- 2)(- k - 1) > 0
9 - 8k - 8 > 0
- 8k + 1 > 0
- 8k > - 1
k < 1/8
So the range of k is k < 1/8

2009-10-04 03:41:55 補充：
Question 4b
concern the largest integral value of k,
we take k = 1/8,
y = - 2x^2 + 3x - 1/8 - 1
y = - 2x^2 + 3x - 9/8
for the x-intercept, sub. y = 0,
- 2x^2 + 3x - 9/8 = 0
16x^2 - 24x + 9 = 0
(4x - 3)^2 = 0
x = 3/4
so the x-intercepts is (3/4, 0)

1a
y = x(1 - x) - k / 4
y = -x^2 + x - k / 4
Because the graph cut x-axis at one point,
-x^2 + x - k / 4 = 0 has one root,
b^2 - 4ac = 0
(1)^2 - 4(-1)(-k / 4) = 0
1 - k = 0
k = 1

1b
the formula is y = -x^2 + x - 1 / 4
for the x-intercept, sub. y = 0,
-x^2 + x - 1 / 4 = 0
4x^2 - 4x + 1 = 0
(2x + 1)^2 = 0
x = -1 / 2
so the coordinates of Q is (-1/2, 0)

2a
y = (x - 4p)(x - 2) + p
y = x^2 - 4px - 2x + 8p + p
y = x^2 + (- 4p - 2)x + 9p
because the graph has one x-intercept,
x^2 + (- 4p - 2)x + 9p = 0 has one root,
b^2 - 4ac =0
(- 4p - 2)^2 - 4(1)(9p) = 0
16p^2 +16p + 4 - 36p = 0
16p^2 - 20p + 4 = 0
4p^2 - 5p + 1 = 0
(4p - 1)(p - 1) =0
p = 1/4 or 1

2b
For p = 1/4,
y = x^2 + [- 4(1/4) - 2]x + 9(1/4)
y = x^2 - 3x + 9/4
for the x-intercept, sub. y = 0,
x^2 - 3x + 9/4 = 0
4x^2 - 12x + 9 = 0
(2x - 3)^2 = 0
x = 3/2
so for p = 1/4, the x-intercept is (3/2, 0)
For p = 1,
y = x^2 + [- 4(1) - 2]x + 9(1)
y = x^2 - 6x + 9
for the x-intercept, sub. y = 0,
x^2 - 6x + 9 = 0
(x - 3) = 0
x = 3
so for p = 1, the x-intercept is (3, 0)
3
y = 9(x - 1)^2 - m - 6
y = 9(x^2 - 2x + 1) - m - 6
y = 9x^2 - 18x + 9 - m - 6
y = 9x^2 - 18x - m + 3
because the graph intersects the x-axis,
b^2 - 4ac >= 0
(- 18)^2 - 4(9)(- m + 3) >= 0
324 + 36m - 108 >= 0
216 + 36m >= 0
36m >= - 216
m >= - 6
so that the smallest value of-6
4a
y = - 2x^2 + 3x - k - 1
because the graph cut the x-axis at 2 points,
b^2 - 4ac > 0
(3)^2 - 4(- 2)(- k - 1) > 0
9 - 8k - 8 > 0
- 8k + 1 > 0
- 8k > - 1
k < 1/8
So the range of k is k < 1/8



4b
concern the largest integral value of k,
we take k = 1/8,
y = - 2x^2 + 3x - 1/8 - 1
y = - 2x^2 + 3x - 9/8
for the x-intercept, sub. y = 0,
- 2x^2 + 3x - 9/8 = 0
16x^2 - 24x + 9 = 0
(4x - 3)^2 = 0
x = 3/4
so the x-intercepts is (3/4, 0)