Expert: Sherman D. - 1/11/2009

Question
1) The first three terms of an arithmetic series are k, 7.5 and k + 7

a) Find the value of k
b) Find the sum of the first 31 terms of this series.

2) f(x) = x² - kx + 9

a) Find the set of values of k foe which the equation f(x) = 0 has no real solutions.
b) Given that k = 4, complete the square (do not solve)
c) Write down the minimum value of f(x) and the value of x for which this occurs

3) The points A and B have co-ordinates (4 , 6) and    (12 , 2) , the straight line L1 passes through A and B
a) Find an equation for L1 in the form ax + by = c
b) The straight line L2 passes through the origin and has gradient -4, write down an equation
c) The lines L1 and L2 intercept at point C, find the exact co-ordinates of the mid-point of AC  

Answer
an = a1 + d(n - 1)

an = k + d(n - 1)

a(2) = k + d(2 - 1)
7.5 = k + d

a(3) = k + d(3 - 1)
k + 7 = k + 2d
2d = 7
d = 3.5

a.)
k + d = 7.5
k + 3.5 = 7.5
k = 4

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b.)

the formula would be a(n) = 4 + (7/2)(n - 1)

S(n) = (n/2)(a1 + a(n))

a(31) = 4 + (7/2)(31 - 1)
a(31) = 4 + (7/2)(30)
a(31) = 4 + 7(15)
a(31) = 4 + 105
a(31) = 109

S(31) = (31/2)(4 + 109)
S(31) = 15(113)
S(31) = 1695

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2.)
f(x) = x^2 - kx + 9

a.)
since f(x) = x^2 + 9 would give you x-intercepts of (i,0) and (-i,0)
ANS : k = 0

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b.)
f(x) = x^2 - 4x + 9

x^2 - 4x + 9 = 0
x^2 - 4x = -9
x^2 - 4x + 4 = -5
(x - 2)^2 = -5
x - 2 = sqrt(-5)
x = 2 ± isqrt(5)

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c.)
If your talking about the vertex, then

x = -b/(2a)
x = -(-4)/(2(1))
x = 4/2
x = 2

f(2) = 2^2 - 4(2) + 9
f(2) = 4 - 8 + 9
f(2) = -4 + 9
f(2) = 5

ANS : (2,5)

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3.)
a.)
(4,6) and (12,2)
m = (2 - 6)/(12 - 4)
m = -4/8
m = (-1/2)

(4,6), m = (-1/2)
6 = (-1/2)(4) + b
6 = -2 + b
b = 8

y = (-1/2)x + 8
2y = -x + 16

ANS : x + 2y = 16

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b.)
(0,0), m = -4
0 = -4(0) + b
0 = 0 + b
b = 0

y = -4x or 4x + y = 0

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c.)
y = -4x
y = (-1/2)x + 8

-4x = (-1/2)x + 8
-8x = -x + 16
-7x = 16
x = (-16/7)

y = -4(-16/7)
y = (64/7)

ANS : ((-16/7),(64/7))