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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

Mathematics
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|dw:1352549825053:dw|
(C) c < 0 as graph cuts y-axis below the x-axis a < 0 because graph is frowning curve
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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Other answers:

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.
|dw:1352551798883:dw|
@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
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.
  • 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
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yes
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yup, that's right.
|dw:1352552484497:dw|
1/(1/x) = x. try again.
oh -.-|dw:1352552602486:dw|
you still need to combine x + x
2x-(2/x)
that's right!
|dw:1352552793347:dw|
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?
|dw:1352553510302:dw|
\[\huge f(-x) = \frac{ (-x) }{ 1-(-x) }\]
errrr.....wait
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))
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
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
I still cannot get the answer
Please start a new post for a new question. If possible, please also put them in the right categories (subtopics). Thanks!
\[\large f(1/x)f(-x) = \frac{ 1/x }{ 1-1/x }\frac{ (-x) }{ 1-(-x) }\] \[\large = \frac{ -1 }{ 1+x -\frac{ 1 }{ x }-1 } \times \frac{ -x }{ -x }\]
so complicated....
just carry out the multiplication and you'll get (D)
|dw:1352556469608:dw| I get this only...
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)
you got it @kryton1212 ???? take it a simply way..,hehe
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|
@hartnn would you kindly help me?
@kryton1212 Please start a new post for a new question!
the problem is I cannot draw picture in a new post
ok, let this be last question discussed in this post, agreed @kryton1212 ? for this u know how to find intercepts ?
You can post the question a draw it in the comment
@hartnn that's ok. c=y-intecept
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 your x-intercept is negative, that is it will cut x axis in negative (left) side got this ? what about y-intercept ???
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...
  • phi
@kryton1212 it would be better if you made each question a separate post.
okok

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