F.4 maths (15 marks)

用戶7600

2010-01-01 7:00 am

Solve the following questions with steps:

1. In a figure, the straight line L: y=ax+b passes through P(2,8). It has slope 2.
(a). Find the values of a and b.
(b). If f(x)=(ax+b)^2, solve f(x)=12.

2. Find the value of k if 2x+ky-8=0 and x+3y+1=0 are the equations of two parallel lines.

3. Let f(x)=x^2-bx+14, where b is a positive constant. V(b/2,-2) is the vertex of the graph of y=f(x).
(a)(1). Find the value of b.
(2). Find the axis of symmetry of the graph of y=f(x).
(b) It is given that g(x)=-f(x+1)+k. The graph of y=g(x) touches the x-axis at N(n,0).
(1). Find the values of k and n.
(2). Solve for x if f(x)+g(x)=0.

回答 (1)

用戶7639

2010-01-01 7:51 am

✔ 最佳答案

1(a)
the straight line L: y=ax+b passes through P(2,8).
8=2a+b__________________________(1)
the straight line L has slope 2
a=2
Subst. a=2 into (1)
8=2(2)+b
b=4

(b)
f(x)=(ax+b)2
=(2x+4)2

f(x)=12
(2x+4)2=12
2x+4=√12 or 2x+4=-√12
2x=-4+√12 or 2x=-4-√12
2x=-4+2√3 or 2x=-4-2√3
x=-2+√3 or x=-2-√3

2.
2x+ky-8=0 and x+3y+1=0 are the equations of two parallel lines.
-2/k=-1/3
-6=-k
k=6

3. (a)(1)
V(b/2,-2) is the vertex of the graph of y=f(x).
-2=(b/2)2 - b(b/2)+14
-2=b2/4 - b2/2+56
-8=b2-2b2+56
-64=-b2
b2=64
b=8 or b=-8(rej.)

(2)
the axis of symmetry of the graph:x=b/2
i.e x=4

(b)(1)
g(x)=-f(x+1)+k.
=-[(x+1)2-b(x+1)+14]+k
=-(x2+2x+1-bx-b+14)+k
=-x2+(b-2)x+(b-15)+k
=-x2+6x+(k-7)

The graph of y=g(x) touches the x-axis at N(n,0),we have
0=-n2+6n+(k-7)________________________(1)
(1) △=0
(6)2-4(-1)(k-7)=0
36+4k-28=0
4k=-8
k=-2

Subst. k=-2 into (1),we have
0=-n2+6n-9
(n-3)2=0
n=3(repeated root)

(2).
f(x)+g(x)=0.
(x2-8x+14)+(-x2+6x-9)=0
-2x+5=0
2x=5
x=5/2

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