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Does the graphical representation of the quadratic function

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03 Oct 2012, 00:20
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Does the graphical representation of the quadratic function f(x) = y = ax^2 + c intersect with the x - axis?

(1) a < 0
(2) c > 0
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03 Oct 2012, 03:53
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Does the graphical representation of the quadratic function f(x) = y = ax^2 + c intersect with the x - axis?

The question basically asks whether $$y$$ can be zero for some value(s) of $$x$$. So, whether $$ax^2+c=0$$ has real roots. $$ax^2+c=0$$ --> $$x^2=-\frac{c}{a}$$ --> this equation will have real roots if $$-\frac{c}{a}\geq{0}$$.

(1) a < 0. Not sufficient since no info about c.
(2) c > 0. Not sufficient since no info about a.

(1)+(2) $$a<0$$ and $$c>0$$ means that $$-\frac{c}{a}=-\frac{positive}{negative}=-negative=positive>{0}$$. Sufficient.

Hope it's clear.
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09 May 2014, 07:11
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qlx wrote:
For the graph to intersect the x axis y = 0
or ax^2 + c =0
for the above equation to be true
x^2 = -c/a

using 1 + 2 we can only tell c/a is negative hence - c/a is positive

but how can we be sure that x^2 = -c/a ? or how can we be sure that ax^2 + c =0

x= 2 , c= 1 and a = -1 here both 1 + 2 is satisfied but ax^2 + c not equal to 0

x= 2 c= 4 and a = -1 here both 1 + 2 is satisfied and ax^2 + c = 0

so we can get both a yes and no depending upon the individual values of C and a it seems,

Is there anything wrong with the above logic

I think you misinterpreted the question. As long as $$-\frac{c}{a}=positive$$, y = ax^2 + c will intersect with the x - axis, because in this case $$ax^2+c=0$$ will have real solution(s).
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03 Oct 2012, 00:50
Does the graphical representation of the quadratic function f(x) = y = ax^2 + c intersect with the x - axis?

1. a <0
2. c >0

Hi,

In order to find whether y=ax^2 + c intersect the x-axis we need to find whether y=0 is possible to achieve for sure. In short if ax^2 + c =0 can be achieved for sure!

STAT1
a <0
now ax^2 will become negative as a is -ve and x^2 is positive
but we dont know the sign of c.
if c<=0 then for sure the curve will not intersect x-axis as ax^2 + c will become negative.
if c>0 then the curve WILL intersect x-axis for sure as ax^2 + c= 0 will give us atleast one solution.

so, NOT SUFFICIENT.

STAT2
c >0
In this case we do not know the sign on a
if a>=0 then we DO Not have a soltuion.
if a <0 then we DO have a solution (as explained above)

So, Not Sufficient.

Combining both we have
a <0 and c>0 and ax^2 + c =0 will give us atleast one soltuion.
So, the curve will intersect x-axis.
So, SUFFICIENT!

Hence, asnwer will be C.
Hope it helps!
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28 Nov 2013, 11:31
The equation ax^2 + c will intersect or touches the x-axis if it has real roots.
Discriminant >= 0. Hence, b^2 - 4ac >=0
i.e. -4ac >=0. We can only predict this when we know about the signs of a and c both. Hence C.

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29 Mar 2014, 09:13
Graph approach also works here

Is y = ax^2 + c?

Well, let's see

Statement 1 a<0

Says parabola* is downward sloping but we don't know about the vertex so could or could not intersect 'x' axis.

Insufficient

Statement 2

C>0

So y intercept is positive but vertex could be above x=0, and thus will not have any solutions for 'x'

Insufficient

Both together

So we know that curve is downward sloping and that 'y' intercept is positive therefore it must at least one solution for 'x' therefore this statement is sufficient

*Just curious, is the line in fact a parabola?

Thanks

Cheers
J

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09 May 2014, 05:48
For the graph to intersect the x axis y = 0
or ax^2 + c =0
for the above equation to be true
x^2 = -c/a

using 1 + 2 we can only tell c/a is negative hence - c/a is positive

but how can we be sure that x^2 = -c/a ? or how can we be sure that ax^2 + c =0

x= 2 , c= 1 and a = -1 here both 1 + 2 is satisfied but ax^2 + c not equal to 0

x= 2 c= 4 and a = -1 here both 1 + 2 is satisfied and ax^2 + c = 0

so we can get both a yes and no depending upon the individual values of C and a it seems,

Is there anything wrong with the above logic

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12 Oct 2014, 06:51
Does the graphical representation of the quadratic equation Y=Ax^2 + C intersect with x-axis?
We are required to find x intercept of equation Y=Ax^2+C
put y=0 to find value of x intercept
0=Ax^2+C
x^2=-C/A
x^2 can't have a negative value. so we need to check whether -C/A has a negative or positive sign.
If x^2 is positive, then the solution is real and if it is negative then solution is unreal.

statement 1 :
A<0
A is negative, but this information is not sufficient to decode on sign of -C/A

statement 2 :
C>0
C is positive, but this information is not sufficient to decode on sign of -C/A

1+2 Combined,
-C/A has a positive value, so solution is real and Y=Ax^2 + C intersect with x-axis.

Ans = C

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12 Oct 2014, 17:09
its simply C.

for x to have have roots we can check the discriminant. sqrt(b^2-4ac). i.e sqrt(b^2-4ac)>=0

here b =0 so the value of sqrt(b^2-4ac) is sqrt(-4ac)

so the question asks us : is sqrt(-4ac)>0 (cannot be zero as per the options)

now u can see the statements on by one:

1)a<0 nt suff

ok A is negative then C can be -ve(imaginary roots) or +ve( then it has roots)

2)C>0 nt suff

ok C is positive then A can be +ve(imaginary roots) or -ve( then it has roots)

taking both statements we conclude sqrt(-4ac)>0

so C.

kudos if u like my review.

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24 Sep 2017, 10:00
Bunuel wrote:
Does the graphical representation of the quadratic function f(x) = y = ax^2 + c intersect with the x - axis?

The question basically asks whether $$y$$ can be zero for some value(s) of $$x$$. So, whether $$ax^2+c=0$$ has real roots. $$ax^2+c=0$$ --> $$x^2=-\frac{c}{a}$$ --> this equation will have real roots if $$-\frac{c}{a}\geq{0}$$.

(1) a < 0. Not sufficient since no info about c.
(2) c > 0. Not sufficient since no info about a.

(1)+(2) $$a<0$$ and $$c>0$$ means that $$-\frac{c}{a}=-\frac{positive}{negative}=-negative=positive>{0}$$. Sufficient.

Hope it's clear.

for ax^2 + c = 0 to have real roots.

b^2 - 4ac >= 0 ; 0^2 - 4ac >= 0 --> ac <= 0

1. a < 0 ; nothing about c ; insufficient
2. c>0 ; nothing about a ; insufficient

1+ 2 ; a & b have opposite signs. hence ac<0

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Re: Does the graphical representation of the quadratic function   [#permalink] 24 Sep 2017, 10:00
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