The graph of y=ax^(2)+bx+c is given as shown. Which of the following is / are true?
I. a<0
II. b<0
III. c<0
A. I only
B. I and II only
C. I and III only
D. II and III only
E. I, II and III

- anonymous

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- anonymous

|dw:1352549825053:dw|

- Kira_Yamato

(C)
c < 0 as graph cuts y-axis below the x-axis
a < 0 because graph is frowning curve

- anonymous

E)
inverted U shape meaning a<0
intersecting y axis at negative means C is less than 0 c<0
and also there is shift along x axis in right direction means b<0

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## More answers

- Kira_Yamato

Overlooked the detail for b<0... Sorry

- anonymous

I am confusing why b>0 is correct

- sirm3d

the line of symmetry x = -b/2a, is a positive value, and we know a < 0 so b > 0

- anonymous

I am sorry that I type it wrongly.
It should be : "I am confusing why b<0 is correct"

- sirm3d

b > 0 is correct.

- anonymous

why

- anonymous

a<0 because parabola opens downward

- anonymous

I know a<0 and c<0, I just don't know why b<0 is correct

- sirm3d

b < 0 is NOT correct.

- anonymous

so the answer is C?

- sirm3d

yes. a < 0 and c < 0.

- anonymous

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- sirm3d

@hashsam1 the line of symmetry is the line x = -b/2a. according to the graph it is to the right of the y-axis (so the sign of -b/2a is positive). we know that a < 0, so if b < 0 then the sign of -b/2a is -(-)/2(-) is negative.

- phi

a simple example with 2 positive roots x=1 and x=2
the polynomial would be -(x-1)(x-2) = -x^2 +3x -2
this suggests a<0 b>0 c<0

- anonymous

I understand now, thank you so much :)

- anonymous

another question
If f(x)=x-(1/x), then f(x)-f(1/x)= ?

- anonymous

oops yes right i over looked the equation thinking its just a normal quadratic equation.

- phi

If f(x)=x-(1/x), then f(x)-f(1/x)= ?
replace f(x) with x- 1/x
replace f(1/x) with 1/x - x

- anonymous

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- sirm3d

yes

- anonymous

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- sirm3d

yup, that's right.

- anonymous

|dw:1352552484497:dw|

- sirm3d

1/(1/x) = x. try again.

- anonymous

oh -.-|dw:1352552602486:dw|

- sirm3d

you still need to combine x + x

- anonymous

2x-(2/x)

- sirm3d

that's right!

- anonymous

|dw:1352552793347:dw|

- sirm3d

i'll accept either form.

- anonymous

another question
If f(x)=10^(2x), then f(4y)=?

- anonymous

A. 10^(4y).
B. 10^(2+4y)
C. 10^(8y)
D. 40^(y)
E. 40^(2y)

- sirm3d

replace x by 4y, what do you get?

- anonymous

B ?

- sirm3d

C it is.

- anonymous

oh yes

- sirm3d

\[\huge 10^{2(4y)}\]

- nubeer

f(x)=10^(2x)
f(x)=10^(2(4y))

- anonymous

If f(x)=x/(1-x), then f(1/x)f(-x)=?

- anonymous

@gerryliyana

- sirm3d

and your answer is?

- anonymous

|dw:1352553510302:dw|

- sirm3d

\[\huge f(-x) = \frac{ (-x) }{ 1-(-x) }\]

- anonymous

errrr.....wait

- anonymous

If f(x)=x/(1-x), then f(1/x)f(-x)=
f(1/x) = (1/x)/(1-(1/x)) = 1/x((x-1)/x)
f(-x) =(-x)/(1-(-x)) = -x/(1+x)
f(1/x)f(-x) = 1/x((x-1)/x) times -x/(1+x)
f(1/x)f(-x) = -x/(x((x-1)/x)(1+x))

- anonymous

o.o errrr.... wait a moment

- anonymous

f(1/x)f(-x) = -x/(x((x-1)/x)(1+x))
f(1/x)f(-x) = -x/((x-1)(x+1))

- anonymous

|dw:1352554094668:dw|

- anonymous

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- sirm3d

|dw:1352554209982:dw|

- anonymous

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- anonymous

is it necessary for you to give me a big cross.......
I got it :)

- anonymous

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- anonymous

|dw:1352554433711:dw|

- anonymous

|dw:1352554512663:dw|

- sirm3d

|dw:1352554536556:dw|

- anonymous

|dw:1352554601402:dw|

- anonymous

|dw:1352554549107:dw|

- sirm3d

|dw:1352554649051:dw|

- anonymous

|dw:1352554682501:dw|

- sirm3d

we're not discussing' we're exchanging notes. hahaha

- anonymous

|dw:1352554679681:dw|

- anonymous

haha :))

- anonymous

|dw:1352554776614:dw|

- anonymous

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- sirm3d

what's the answer to your problem \[\huge f(\frac{ 1 }{ x })f(-x)\]

- anonymous

oh my god

- anonymous

A. -1/2
B. -1
C. -(1-x)/(1+x)
D. x/(1-x^(2))
E. x/(x^(2)-1)
no answer since this is past paper, my teacher haven't give us the answers yet

- anonymous

I still cannot get the answer

- Callisto

Please start a new post for a new question. If possible, please also put them in the right categories (subtopics). Thanks!

- sirm3d

\[\large f(1/x)f(-x) = \frac{ 1/x }{ 1-1/x }\frac{ (-x) }{ 1-(-x) }\] \[\large = \frac{ -1 }{ 1+x -\frac{ 1 }{ x }-1 } \times \frac{ -x }{ -x }\]

- anonymous

so complicated....

- sirm3d

just carry out the multiplication and you'll get (D)

- anonymous

|dw:1352556469608:dw|
I get this only...

- anonymous

f(1/x)f(-x) = -x/(x((x-1)/x)(1+x))
f(1/x)f(-x) = -x/((x-1)(x+1))
f(1/x)f(-x) = -x/(x^2 -1)
f(1/x)f(-x) = -x/-(-x^2 +1)
f(1/x)f(-x) = x/(1-x^2)
the answer is D x/(1-x^2)

- anonymous

you got it @kryton1212 ???? take it a simply way..,hehe

- anonymous

confusing................. I think it later
another question
If a,b and c are all positive, which of the following may represent the graph of ax+by+c=0?|dw:1352556954406:dw|

- anonymous

@hartnn would you kindly help me?

- Callisto

@kryton1212 Please start a new post for a new question!

- anonymous

the problem is I cannot draw picture in a new post

- hartnn

ok, let this be last question discussed in this post, agreed @kryton1212 ?
for this
u know how to find intercepts ?

- Callisto

You can post the question a draw it in the comment

- anonymous

@hartnn that's ok. c=y-intecept

- hartnn

nopes, c is y intercept in y=mx+c
not here

- hartnn

to get x intercept, put y=0 in ax+by+c=0
to get y intercept, put x=0 in ax+by+c=0

- hartnn

now tell me what intercepts u get

- anonymous

....oh yes.
I don't know how to find the intercept in these gaphs

- hartnn

to get x intercept, put y=0
to get y intercept, put x=0
in any graphs

- anonymous

ax+by+c=0
put y=0???
ax+c=0?

- hartnn

yes, thats right
ax+c=0
x= ?

- anonymous

x=-c/a?

- hartnn

absolutely correct!
now -c/a is positive or negative ?

- hartnn

given a and c are positive....

- anonymous

-ve

- hartnn

so your x-intercept is negative, that is it will cut x axis in negative (left) side
got this ?
what about y-intercept ???

- anonymous

so it may be B or E.
y=-c/b
is it E?

- hartnn

yes, y intercept is also negative, so it will cut down.
and yes its E
clear as to how ?

- anonymous

clear now, thank you

- hartnn

make sure u ask next question in new post...

- phi

@kryton1212 it would be better if you made each question a separate post.

- anonymous

okok

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