F.4 quadratic Functi0ns~~

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

(b)(i)
c=6

(b)(ii)
Minimum value of the function
=[4(2)(6)-4^2]/(4*2)
=5

2010-03-03 19:14:53 補充：
Some notes for your revision:
for a quadratic function y=ax^2+bx+c, where a, b, c are real constants and a is not equal to 0,
x-coordinate of vertex of graph = -b/(2a)
hence axis of symmetry of graph is x = -b/(2a)
y-coordinate of vertex of graph = (4ac-b^2)/(4a)
y-intercept of graph = c

2010-03-03 19:16:05 補充：
Also,
if a>0,
(4ac-b^2)/(4a) = minimum value of graph
if a<0,
(4ac-b^2)/(4a) = maximum value of graph

2010-03-03 19:58:39 補充：
sorry, for part (b)(ii), there is a typing error of the final answer
the answer should be 4

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

b)i)when x=0,y=6
2(0)^2+4(0)+c=6
c=6

ii)y=2x^2+4x+6
=2(x+1)^2+4
because the minium value of (x+1)^2 is 0
the minium of y is
=2(0)+4
=4

2010-03-03 19:46:31 補充：
系第1位回答者

It is given that the axis of symmetry of the graph of the function .

y=2x^2+bx+c is x =－1.

(a) Find the value of the constant b.

Sol.

Set y=2(x+1)^2+p.

y=2(x^2+2x+1)+p=2x^2+4x+p+2.

b=4

(b) If the graph cuts the y-axis at (0，6)，find

(i) the value of the constant c

y=2x^2+4x+c.

6=0－0+c.

c=6.

(iI) the minimum value of the function.

y=2x^2+4x+6.

=2(x^2+2x+1)+4.

=2(x+1)^2+4.

(x+1)^2>=0.

2(x+1)^2+4>=4.

y>=4.

我好唔明啊 可唔可以解一解=__=

2010-03-03 19:26:06 補充：
f.4正係學左quadratic formula
同completing square method...
未學你上述講既formula~
請問有無其他方法搵呢?
唔該哂!

2010-03-03 20:12:16 補充：
(a) 點解有個p 咩野來-__-

=[4(2)(6)-4^2]/(4*2)
=4