Maths Questions I can't seem to solve :(

Deadly SG

These 2 questions I have been pondering over for some time now, can't seem to be able to find a solution to them. Need help!

1) A quadratic function f(x) = k(x - 1)(x - m) is such that m > 1. The point (0, -12) lies on hte graph of y = f(x).

a) Write down an equation connecting k and m.
b) Write down the coordinates of the points where the graph cuts the x-axis.
Given that the function has a maximum value of 6.75,
c) Write down the coordinates of the maximum point of y = f(x) in terms of m.
d) Hence, find the value of m.

2) A polynomial P(x) is such that P(x) = px^4 + 4x^3 - 3x^2 + qx - 5. When P(x) is divided by x^2 - 1 the remainder is 5x - 6.

a) Express P(x) in terms of x^2 - 1 and the remainder 5x - 6. Hence find the values of p and q.
b) Solve the equation P(x) = 2x^4 + 7x - 10.

Thanks in advance Very Happy

eagle SG

a) sub x = 0
-12 = km

b) immediately, u should see that k is negative
and graph cuts x-axis at 1 and m

c)
centre is at x=(1+m)/2 which will give you the value of 6.75
hence, ans
( {1+m}/2 , 6.75)

d) you can differentiate and find the value of m from here.

eagle SG

a) P(x) = Q(x) (x^2-1) + 5x - 6
=> Not sure if the form is like that

sub x = 1
P(1) = p + 4 - 3 + q - 5 = 5 - 6
hence, p+q = 3

sub x = -1
P(-1) = p - 4 - 3 - q - 5 = -5 - 6
hence, p-q = 1

solving, p=2, q=1

b) I blur at the moment... See again later... Very tired

Deadly SG

Originally posted by eagle:
c)
centre is at x=(1+m)/2 which will give you the value of 6.75
hence, ans
( {1+m}/2 , 6.75)[/b]

edit: my mistake

hisoka SG

part b. let px = (ax^2 + bx +c) * (x^-1) + 5x - 6

then expand and compare coeff. on both sides. should give you 5 eqns and 5 unknowns solve for p and q.

^tamago^ SG

1a) since (0,-12) is a point on f(x), when x=0, f(x)=-12=k(-1)(-m)=km km=-12

b) when f(x) cuts x-axis, y=0, i.e. k(x-1)(x-m)=0
x=1 or m
(1,0), (m,0)

c) when f(x) is max, f(x)=6.75 and k1
(½(m+1),6.75)

d) when x=½(m+1), f(x)=k[½(m+1)-1][½(m+1)-m]=6.75
k(m-1)(1-m)=27 [multiply by 4]
(12/m)(m-1)²=27 [k=(12/m)(-1)]
4(m-1)²=9m [multiply by m/3]
4m²-17m+4=0
(4m-1)(m-4)=0
m=4 or ¼ (rej. as m>1)

^tamago^ SG

2a) P(x) = px^4 + 4x³ - 3x² + qx - 5
P(x) = (x²-1)Q(x) + 5x-6 for some Q(x) is a quadratic eqn
when x=1, P(1) = p+4-3+q-5 = 5(1)-6 = -1 -----> p+q=3 ---(1)
when x=-1, P(-1) = p-4-3-q-5 =5(-1)-6 = -11 -----> p-q=1 ---(2)
(1)+(2): p=2
sub p=2 into (1): q=1

b) P(x) = 2x^4 + 7x - 10
x=1.06332 or -1.83896
[Pls check, answers are non-integer]

tanjun SG

For part b),

Taking answer from a) which is p = 2, q = 1,

P(x) = 2x^4 + 7x - 10
2x^4 + 4x^3 - 3x^2 + x - 5 = 2x^4 + 7x - 10
4x^3 - 3x^2 - 6x + 5 = 0

Let h(x) = 4x^3 - 3x^2 - 6x + 5
h(1) = 4 - 3 - 6 + 5
= 0

Thus, x - 1 is a factor of h(x)

You can use long division to find the quotient.

However, I like to use another method called synthetic division.

4`````-3```````-6````````5

``````4````````1````````-5
```````````````````````````````````1
_______________________________
``````4````````1````````-5````0

From here,

h(x) = (x - 1)(4x^2 + x - 5)
(x - 1)(4x^2 + x - 5) = 0
(x - 1)(4x + 5)(x - 1) = 0
Therefore, x = 1 (repeated) or x = -5/4