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

Overlooked the detail for b<0... Sorry

I am confusing why b>0 is correct

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

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

b > 0 is correct.

why

a<0 because parabola opens downward

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

b < 0 is NOT correct.

so the answer is C?

yes. a < 0 and c < 0.

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I understand now, thank you so much :)

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

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

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

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yes

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yup, that's right.

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

oh -.-|dw:1352552602486:dw|

you still need to combine x + x

2x-(2/x)

that's right!

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i'll accept either form.

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

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

replace x by 4y, what do you get?

B ?

C it is.

oh yes

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

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

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

and your answer is?

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

errrr.....wait

o.o errrr.... wait a moment

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

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

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

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haha :))

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

oh my god

I still cannot get the answer

so complicated....

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

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I get this only...

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

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

the problem is I cannot draw picture in a new post

You can post the question a draw it in the comment

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

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

now tell me what intercepts u get

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

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

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

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

x=-c/a?

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

given a and c are positive....

-ve

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

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

clear now, thank you

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

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

okok