Problem on quadratic polynomial.

[sankalpmittal]

Homework Statement

The graph of the quadratic polynomial , y=ax²+bx+c is as shown below in the figure :

http://postimage.org/image/nvkxv74yd/

Then :

(A) b²-4ac<0
(B) c<0
(C) a<0
(D) b<0

Homework Equations

y=ax²+bx+c
If y=0 , then ax²+bx+c=0
Then ,
x = {-b+-sqrt(b^2-4ac)}/2a

The Attempt at a Solution

Hmm tally image please...
When parabola is intersecting x-axis then y=0 and
Case 1 : x<0
Case 2 : x>0
if x >0 or x<0
then b²-4ac >0
So b² > 4ac

Now what ...

Please help !

Thanks in advance.

 

[quantumdude]

You're thinking too hard. Think about the concavity of the graph (that is, the direction in which the parabola opens).

 

[sankalpmittal]

You're thinking too hard. Think about the concavity of the graph (that is, the direction in which the parabola opens).

Thanks for a hint ex-mentor !
Parabola is opening in the -y axis direction , right ?

 

[Morgoth]

If A was correct:
it would mean that y=ax^2+bx+c =0 would have no real solution, which would mean that the y(x) would never be zero, which is wrong by the graph.
Wrong

If B was correct:
it would mean that for x=0 the y(x)=y(0) would take the value c <0 which by the graph is incorrect. y(0)>0.
Wrong

Now C :
You check y(x) derivative.
y'(x)= 2ax+b.
and from that we see that y'(0)= b.
Though since your MAXIMUM is at x=0, it means that y'(x) must be zero at that point y'(0)=0.
Wrong

Checking on D.
The way to see it is to look at its second derivative:
y''(x)= 2a.
you see from the graph that a has to be a<0 because of the concavity.
Right

 

[sankalpmittal]

If A was correct:
it would mean that y=ax^2+bx+c =0 would have no real solution, which would mean that the y(x) would never be zero, which is wrong by the graph.
Wrong

If B was correct:
it would mean that for x=0 the y(x)=y(0) would take the value c <0 which by the graph is incorrect. y(0)>0.
Wrong

Now C :
You check y(x) derivative.
y'(x)= 2ax+b.
and from that we see that y'(0)= b.
Though since your MAXIMUM is at x=0, it means that y'(x) must be zero at that point y'(0)=0.
Wrong

Checking on D.
The way to see it is to look at its second derivative:
y''(x)= 2a.
you see from the graph that a has to be a<0 because of the concavity.
Right

Please don't go so sophisticated. We haven't been taught limits and derivatives and integrals till now because I am after all in class 10th , 15 years. However your answer is correct and that was only given in answer key. Can it be done without involving introductory calculus ?

You are a new member. Please don't give full answer as its against forum rules.
Just alerting you.

 

[Morgoth]

Ups sorry...I didn't know...I will keep that in mind.

Oh well at least for the 2 firsts i think you don't need calculus.
Now the rests, you can check anytime.
Try drawing for example:
y=ax^2+bx+c

with a,b,c so that you will have your cases.

eg for case A it is y=x^2 + x + 1 b^2-4ac= -3 <0

and see how it works :) then see which one fits better

 

[I like Serena]

Hi sankalpmittal!

Let's try and categorize your drawing of the parabola a bit.

How many roots do you have?
And what consequence does that have for your solution formula?

At which x coordinate do you have the top of the parabola?
And at which side of the x-axis is the top?
Can you substitute that in the generic parabola equation and draw some conclusions?

You already saw it opens in the negative y-direction.
What happens if you substitute a very large x-value in the parabola equation?
What would the sign be of the resulting y-value?

 

[quantumdude]

Thanks for a hint ex-mentor !
Parabola is opening in the -y axis direction , right ?

Yes that's right. What, if anything, does that imply for the coefficients a, b, or c?

 

[NascentOxygen]

When x is very large, y is approx equal to ax²
Since x² is going to be positive, the sign of y in this region tells you something about a.

 

[sankalpmittal]

Many thanks to NascentOxygen , Tom Mattson , ILS and Morgoth for their hints !

I think I figured out this problem.

The first option A i.e. b²-4ac<0 is incorrect because that parabola is yielding real roots. Option A is out.

When that parabola is cutting the y-axis then of course x=0
So
this implies that
y=a.0²+b.0+c
y=c

As we know from graph that y is positive so c is also positive. This implies that c>0.
But option B says that c<0 which is wrong. Option B is out.

Now when parabola is intersecting on first quadrant (+) of x-axis , then y=0 (we can also take -x axis also.)
so
0=ax²+bx+c
As we know that
m = Δy/Δx
We know from above equation
m = (y-c)/x where c is y intercept. As y = 0 then also c =0 because Δy=0 here. We aren't taking in account y intercept because its 0 !

This implies y=c=0

So we know b²-4ac>0 as its yielding real roots.
so b²-4.a.0>0
b²>0
or₁
b>0

But option D says that b<0 which is incorrect as we found that b>0 so option D is out.

Now only option which is left is C which has to be correct because ,
x = {-b+-sqrt(b^2-4ac)}/2a
as c=0

I get 2 values :

x= (-b+b)/2a
x=0

And
x= -2b/2a
x=-b/a
Now here x is positive because axis is +x. To make "x" positive "a" has to be negative as we know before that "b" is positive. This implies that a<0.
QED.

I got the correct answer but am I methodically correct ?

Thanks again !

 

[I like Serena]

The first option A i.e. b²-4ac<0 is incorrect because that parabola is yielding real roots. Option A is out.

When that parabola is cutting the y-axis then of course x=0
So
this implies that
y=a.0²+b.0+c
y=c

As we know from graph that y is positive so c is also positive. This implies that c>0.
But option B says that c<0 which is wrong. Option B is out.

Yep and yep! :)

So c is the y-intercept of the parabola.

Now when parabola is intersecting on first quadrant (+) of x-axis , then y=0 (we can also take -x axis also.)
so
0=ax²+bx+c
As we know that
m = Δy/Δx
We know from above equation
m = (y-c)/x where c is y intercept. As y = 0 then also c =0 because Δy=0 here. We aren't taking in account y intercept because its 0 !

This implies y=c=0

Huh?
I do not understand.
Where did m = Δy/Δx come from?
And why would m = (y-c)/x where c is y intercept?

The tangent line would not have the same y-intercept as the parabola has...

c is not zero.
You already concluded it was greater than zero and that is was the y-intercept of the parabola...

 

[sankalpmittal]

Huh?
I do not understand.
Where did m = Δy/Δx come from?
And why would m = (y-c)/x where c is y intercept?

The tangent line would not have the same y-intercept as the parabola has...

c is not zero.
You already concluded it was greater than zero and that is was the y-intercept of the parabola...

I already concluded that c=0 because the parabola is not cutting y-axis when y=0 and x>0.

If I am wrong then what shall I do ? Any more hints...
Up to now I have found that option A and B are out.

Hmm so
0=ax²+bx+c
where b>0 and c>0
Now what..

Also b²-4ac >0

 

[I like Serena]

I already concluded that c=0 because the parabola is not cutting y-axis when y=0 and x>0.

If I am wrong then what shall I do ? Any more hints...
Up to now I have found that option A and B are out.

Hmm so
0=ax²+bx+c
where b>0 and c>0
Now what..

Also b²-4ac >0

You can't get more information from y=0 than you already have.
That is, that there are 2 solutions for y=0, which is expressed in b²-4ac >0.

Also, you can not conclude that b>0.


Let's inventorize:
1. the parabola intersects the positive y-axis at y=c, meaning c>0
2. the parabola intersects the x-axis at 2 points, meaning b²-4ac >0.
3. the parabola opens up downward, meaning...?
4. at which x-coordinate does the parabola have its maximum? And what to deduce?

 

[NascentOxygen]

I already concluded that c=0 because the parabola is not cutting y-axis when y=0 and x>0.

Earlier you determined that c>0, and you did that correctly. So how can you now take another view, and say c=0?

Better go back and refresh your mind on the method that led to you getting it right.

y=ax²+bx+c

 

[sankalpmittal]

Let's inventorize:
1. the parabola intersects the positive y-axis at y=c, meaning c>0
2. the parabola intersects the x-axis at 2 points, meaning b²-4ac >0.
3. the parabola opens up downward, meaning...?
4. at which x-coordinate does the parabola have its maximum? And what to deduce?

Earlier you determined that c>0, and you did that correctly. So how can you now take another view, and say c=0?

Better go back and refresh your mind on the method that led to you getting it right.

Hmmm let me think a bit.

This question was for the standard of class 10th. I can't involve introductory calculus or parabolic equation ( c² = x²+y² ) here. I have to do it by applying simple coordinate geometry (up to 10th standard) only.

Anyways , when parabola is intersecting on positive quadrant of x-axis then ,

x = {-b+-sqrt(b^2-4ac)}/2a , but here I can't say that y=c>0 also. I have got myself totally confused and winded. If I am taking in account that option B that is c<0 wrong since c>0 and also y>0 then I can also take that in account here. I have got myself totally blundered now !

The concavity of parabola is downward meaning that y is negative and at one side x > 0 and y<0 AND at other side y<0 and x<0.

Now what can I do ?

 

[I like Serena]

Suppose we pick the parabola y=2x².
It goes up doesn't it?

What would it take for the parabola to go down?

(I like your picture btw. )

 

[NascentOxygen]

Hmmm let me think a bit.
<neat pic>

It's fortunate that this image is not in the list of available smileys, or writers here would find it applicable to almost every post on PF!

Anyways , when parabola is intersecting on positive quadrant of x-axis then ,

A quadrant is an area, one-quarter of the plane. You could phrase this better as, "where parabola intersects the positive x-axis then ...".

Now what can I do ?

You could start by returning to my question which so far has not been answered:

For the equation y=ax²+bx+c, when x is very large, this becomes y is approx equal to ax²
Since x² is going to be positive, yet in this region on the graph we see that y is negative, then the sign of y in this region tells you something about the sign of a.

 

[ehild]

You do overcomplicate the problem. Look at NascentOxygen's big formula:

y=ax²+bx+c

and compare with your drawing:
The curve intersects the y-axis at a positive y value.The points on the y-axis correspond to x=0. Therefore

c>0

A general parabola can be obtained from the standard y=X² one by shifting it along x, y and multiplying by a number a as

y=a^2+bx+c=a\left((x+\frac{b}{2a})^2+(c/a-\frac{b^2}{4a^2})\right)

If a is a negative number, the parabola is upside down with respect to the standard one. The standard one is open upward. Your parabola is open downward, so
a<0

b/2a shows the shift of the parable along the x direction. As I see, your parabola is not shifted.

ehild

 

[sankalpmittal]

Suppose we pick the parabola y=2x².
It goes up doesn't it?

What would it take for the parabola to go down?

Concavity of parabola is in the direction of -y axis. In y=2x² , x² is positive and so is 2 , so this equation is not satisfied if we compare it with the parabola going downwards as y is negative. You are asking me to use parabolic equation... y=ax² where x² is positive and y is negative , so "a" has to be negative.

This implies a<0 , isn't it ?

(I like your picture btw. )

Thanks !

You could start by returning to my question which so far has not been answered:

Answered , I think. See the above reply to ILS' post in my post.

You do overcomplicate the problem. Look at NascentOxygen's big formula:

y=ax²+bx+c

and compare with your drawing:
The curve intersects the y-axis at a positive y value.The points on the y-axis correspond to x=0. Therefore

c>0

A general parabola can be obtained from the standard y=X² one by shifting it along x, y and multiplying by a number a as

y=a^2+bx+c=a\left((x+\frac{b}{2a})^2+(c/a-\frac{b^2}{4a^2})\right)

If a is a negative number, the parabola is upside down with respect to the standard one. The standard one is open upward. Your parabola is open downward, so
a<0

b/2a shows the shift of the parable along the x direction. As I see, your parabola is not shifted.

ehild

Ahh ! One more easier method ! Parabola is yielding real and distinct roots. So one value of x is positive and another is of course negative. So product of the two roots is negative as well.

P or Product of two roots = c/a
P = c/a
Since "P" is negative and c>0 so of course a<0 ! I couldn't help laughing !

Can this problem be solved without using parabolic equation ?
BTW , how can I say that b<0 i.e. option D is wrong ?

Thanks in advance.

 

[ehild]

Sankalpmittal,
This is a little exercise for you, to get the feeling about quadratic equations and parabolas. Look at the following quadratic functions and sketch them.
What is the product of the roots and the sum of the roots of the f(x)=0 quadratic equations? Find a, b, c for all of them. Find the position of the peaks. Discover how the position of the peak is related to the roots.

f(x)=x²;
f(x)=-x²;

f(x)=(x-2)²;
f(x)=(x+2)²;

f(x)=(x-2)(x+2);
f(x)=(2-x)(x+2);

f(x)=(x-1)(x-3);
f(x)=(1-x)(x-3);

f(x)=(x-1)(x+3);
f(x)=(x+1)((x-3);
f(x)=(3-x)(x+1).

 

[I like Serena]

Concavity of parabola is in the direction of -y axis. In y=2x² , x² is positive and so is 2 , so this equation is not satisfied if we compare it with the parabola going downwards as y is negative. You are asking me to use parabolic equation... y=ax² where x² is positive and y is negative , so "a" has to be negative.

This implies a<0 , isn't it ?

(Snip)

Yep!

You seem to be picking things up nicely! ;)

Can this problem be solved without using parabolic equation ?
BTW , how can I say that b<0 i.e. option D is wrong ?

Erm... all the answers are related to the parabolic equation...
So how would you deduce an answer without it?
I think the trick is to understand what the parameters mean geometrically.
Ehild's exercises should help with that.

And suppose you had simply y=bx+c.
What does it mean if either b<0, b=0, or b>0?
What happens if you add the extra term ax²?

 

[Morgoth]

"Can this problem be solved without using parabolic equation ?
BTW , how can I say that b<0 i.e. option D is wrong ?"

from your solutions x1,x2 you can see that b<0 would mean that:

x1= -b/2a + sqrt[b^2-4ac]/2a
x2= -b/2a - sqrt[b^2-4ac]/2a

[x1+x2]= -b/a => [x1+x2]/2 = -b/2a
so where would the mean value of your solutions (y(x)=0) lie? on which side of the x-axis acording to b and a signs?

Hope this can help a little bit more.

 

[sankalpmittal]

Note : f(x) = 0 for all equations given :
So here ax²+bx+c = 0

1. f(x)=x²;

Product of roots = c/a ; Sum of roots = -b/a
x=0
a=1 , b=0 , c=0
Sum of roots = product of roots = 0

2. f(x)=-x²;

x=0
a=-1 , b=0 , c=0
Sum of roots = product of roots = 0

3. f(x)=(x-2)²;

x=2
a=1 , b=-4 , c=4
Sum of roots = Product of roots = 4

4. f(x)=(x+2)²;

x=-2
a=1 , b=4 , c=4
Sum of roots = -4
Product of roots = 4

5. f(x)=(x-2)(x+2);

x= 2 , -2
a=1 , b=0 , c=-4
Sum of roots =0
Product of roots = -4

6. f(x)=(2-x)(x+2);

x= 2 , -2
a=-1 , b=0 , c=4
Sum of roots =0
Product of roots = -4

7. f(x)=(x-1)(x-3);

x= 1 , 3
a=1 , b=-4 , c=3
Sum of roots =4
Product of roots = 3

8. f(x)=(1-x)(x-3);

x= 1 , 3
a=-1 , b=4 , c=-3
Sum of roots =4
Product of roots = 3

9. f(x)=(x-1)(x+3);

x= 1 , -3
a=1 , b=2 , c=-3
Sum of roots =-2
Product of roots =-3

10. f(x)=(x+1)(x-3);

x= -1 , 3
a=1 , b=-2 , c=-3
Sum of roots =2
Product of roots = -3

11. f(x)=(3-x)(x+1).

x= -1 , 3
a=-1 , b=2 , c=3
Sum of roots =2
Product of roots = -3

Now I am taking your 10th quadratic equation to plot the graph which is attached with the post , a bit too untidy:
f(x)=(x+1)(x-3)
f(x) = x²-2x-3
Let f(x) = y ,
y=x²-2x-3

Wow ! I see that in my graph concavity is in upward direction and so a>0 , b<0 and c<0 !

Yep!

You seem to be picking things up nicely! ;)

Really ?!

Erm... all the answers are related to the parabolic equation...
So how would you deduce an answer without it?
I think the trick is to understand what the parameters mean geometrically.
Ehild's exercises should help with that.

And suppose you had simply y=bx+c.
What does it mean if either b<0, b=0, or b>0?
What happens if you add the extra term ax²?

if b<0 then
then x can be negative or positive and c>0.
y-c = Y is negative , so
Y=ax²+bx
then bx is positive or negative. If positive then a<0 , if negative then...
If b=0 then
Y=ax²
a<0
If b>0
then
Y=ax²+bx
then again same problem arises as in case b<0 !

How about this method :
P or Product of two roots = c/a
P = c/a
Since "P" is negative and c>0 so of course a<0 !

from your solutions x1,x2 you can see that b<0 would mean that:

x1= -b/2a + sqrt[b^2-4ac]/2a
x2= -b/2a - sqrt[b^2-4ac]/2a

[x1+x2]= -b/a => [x1+x2]/2 = -b/2a
so where would the mean value of your solutions (y(x)=0) lie? on which side of the x-axis acording to b and a signs?

Hope this can help a little bit more.

Mean value lies in +x axis according to parabola in post 1.
[x1+x2]/2>0
So -b/2a >0
b>0
if we solve like
1/2a >0
then it would be like a>0/0 which is indeterminate.
So b>0
Now ,
-b/2a >0
Now this implies a<0 !

Am I correct ?

 

[Morgoth]

no you are not correct.
when you said b>0 and said 1/2a>0 why did you then type a>0/0? And don't forget the inequality properties. What happens to the inequality sign when you multiply both sides with a (-).

(-1)*b/2a = -b/2a>0 means that b/2a<0 (two negative numbers [-1 and b/2a<0] multiplied will give you a positive one).

so if a>0, b<0.
if a<0, b>0.
Otherwise if they had the same sign you would have \frac{+}{+}= + >0 or \frac{-}{-}= + >0

And you cannot use what I typed above in a diagram that doesn't have the x-axis calibrated correctly. I just said that for one more calculus way to find the sign of b, knowing the one of a, and not to answer your question but try to give you one additional help of understanding.

 

[sankalpmittal]

no you are not correct.
when you said b>0 and said 1/2a>0 why did you then type a>0/0? And don't forget the inequality properties. What happens to the inequality sign when you multiply both sides with a (-).

(-1)*b/2a = -b/2a>0 means that b/2a<0 (two negative numbers [-1 and b/2a<0] multiplied will give you a positive one).

so if a>0, b<0.
if a<0, b>0.
Otherwise if they had the same sign you would have \frac{+}{+}= + >0 or \frac{-}{-}= + >0

And you cannot use what I typed above in a diagram that doesn't have the x-axis calibrated correctly. I just said that for one more calculus way to find the sign of b, knowing the one of a, and not to answer your question but try to give you one additional help of understanding.

Oops !
My method is wrong but not the result.
I know [x1+x2]/2>0
So -b/2a >0
This means that -b/2a as a whole is a positive value , isn't it. So if I square on both sides then signs will obviously not change !
(-b/2a)² >(0)²
b²/4a² > 0
So b²>0 => b>0
Hence in -b/2a >0
a < 0 !
Your method is easy I see.

 

[Morgoth]

better than squaring- when you have something like:

-(something)> (else)
do:
(-1)*(-something)<(-1)*(else)
so (something)<-(else)

if (else)=0 and something here will be (something)=b/2a
you have:
-b/2a>0
means
b/2a<0

you have understood how to find a's sign so you know that a<0 or a>0 so respectively b>0 or b<0

 

[ehild]

Now I am taking your 10th quadratic equation to plot the graph which is attached with the post , a bit too untidy:
f(x)=(x+1)(x-3)
f(x) = x²-2x-3
Let f(x) = y ,
y=x²-2x-3

Wow ! I see that in my graph concavity is in upward direction and so a>0 , b<0 and c<0 !

Really ?!

Why are you surprised?

I send you a bunch of parabolas to get familiar with them. Notice that they are symmetric, the axis of symmetry is parallel to the y-axis and goes through the peak, and the peak is halfway between the roots. Notice that the concavity depends on the sign of the multiplier of x² alone.
Which parabola is most similar to the one in your original post?

ehild

 

[sankalpmittal]

better than squaring- when you have something like:

-(something)> (else)
do:
(-1)*(-something)<(-1)*(else)
so (something)<-(else)

if (else)=0 and something here will be (something)=b/2a
you have:
-b/2a>0
means
b/2a<0

you have understood how to find a's sign so you know that a<0 or a>0 so respectively b>0 or b<0

Morgoth , I know rules of inequalities. By squaring on both sides I can easily get that b>0 and a<0. If I mess up with - sign here then condition become not so obvious : either a>0 and b<0 , OR a<0 and b>0.

Why are you surprised?

I send you a bunch of parabolas to get familiar with them. Notice that they are symmetric, the axis of symmetry is parallel to the y-axis and goes through the peak, and the peak is halfway between the roots. Notice that the concavity depends on the sign of the multiplier of x² alone.
Which parabola is most similar to the one in your original post?

ehild

I guess that the parabola in somewhat light blue colour is the approximate match of my parabola in post 1 : https://www.physicsforums.com/attachment.php?attachmentid=42155&d=1324617111

 

[Morgoth]

Not quite right. Squaring is going to mix up everything in your mind, because even if you square -1 you will get +1, and then will you go tell that -1>0 because its square is (+1)>0?

Because that's what you do when you type me:"So b²>0 => b>0" in your previous post.b² >0 (for b in ℝ) means that b<0 or b>0 but surely not 0.

So be careful when squaring. Another thing is that you have:
+4 > -6
squaring:
16 > 36 ?Another example:
-1 < 0
squaring:
1<0 !and one last for another reason:
let's say you have one number A and one number B, for which you know that their squares will also follow the below rule:
A² > B²

What you can say for A and B?
the above rule tells you that:
A>B
or
A<-B

for example on this is to try find which number x is:
x² > 2
the answer is that:
x<-√2 or x>+√2

x²<2
gives
-√2 < x < +√2
even when you have equalities it is being difficult.
If you have tha a=b then you can say that a²=b² here without a problem (as there was with -1<0 above)
On the other hand
if you have that a²=b² then you write that a= ±b.

 

[ehild]

I guess that the parabola in somewhat light blue colour is the approximate match of my parabola in post 1 : https://www.physicsforums.com/attachment.php?attachmentid=42155&d=1324617111

Your parabola in post #1 is not a real parabola but it has maximum at x=0. So it must be symmetric to the y axis. It would be useful to see the original picture or a better copy of the original.
If the maximum is really at x=0, what is b then?

ehild

 

[sankalpmittal]

Not quite right. Squaring is going to mix up everything in your mind, because even if you square -1 you will get +1, and then will you go tell that -1>0 because its square is (+1)>0?

Because that's what you do when you type me:"So b²>0 => b>0" in your previous post.b² >0 (for b in ℝ) means that b<0 or b>0 but surely not 0.

So be careful when squaring. Another thing is that you have:
+4 > -6
squaring:
16 > 36 ?Another example:
-1 < 0
squaring:
1<0 !and one last for another reason:
let's say you have one number A and one number B, for which you know that their squares will also follow the below rule:
A² > B²

What you can say for A and B?
the above rule tells you that:
A>B
or
A<-B

for example on this is to try find which number x is:
x² > 2
the answer is that:
x<-√2 or x>+√2

x²<2
gives
-√2 < x < +√2
even when you have equalities it is being difficult.
If you have tha a=b then you can say that a²=b² here without a problem (as there was with -1<0 above)
On the other hand
if you have that a²=b² then you write that a= ±b.

Ok , I was fully messed up until I discovered the correct answer by a correct method using your logic only !

Product of two roots or c/a< 0
And mean of two roots or -b/2a>0
Now on dividing the two I get :
i.e. c/a/-b/2a <0
2c/-b < 0 because +/- gives - , its obvious !
now b has to positive ! So b>0. As c/a <0 here you can see that a<0.

Your parabola in post #1 is not a real parabola but it has maximum at x=0. So it must be symmetric to the y axis. It would be useful to see the original picture or a better copy of the original.
If the maximum is really at x=0, what is b then?

ehild

If the maximum is really at x=0, then b>0 is what I concluded by my discovery on this aspect.

 

[ehild]

If the maximum is really at x=0, then b>0 is what I concluded by my discovery on this aspect.

The maximum or minimum of a quadratic function is at x=-b/2a. (Why?) If it is at x=0, b=0

ehild

 

[Morgoth]

I shall repeat only one thing:
The mean value you can only check if your axis is CALIBRATED. in a general axis where you don't know if x1,x2 you cannot use it.
if x1=-4 and x2=2 then the mean value is at -1<0.
if x1=-1 and x2=5 then the meave value is at +3>0.
when you don't know or have no clue where is each of the two, you CANNOT use it.

For example in your graph i get that the mean value is positive, but i cannot be sure because it is not so obvious.

 

[sankalpmittal]

The maximum or minimum of a quadratic function is at x=-b/2a. (Why?) If it is at x=0, b=0

ehild

Are you taking in account mean value of x ? If yes , then mean value of x is 0 as x is itself 0. Mean of x=-b/2a => 0=-b/2a => b=0.

I shall repeat only one thing:
The mean value you can only check if your axis is CALIBRATED. in a general axis where you don't know if x1,x2 you cannot use it.
if x1=-4 and x2=2 then the mean value is at -1<0.
if x1=-1 and x2=5 then the meave value is at +3>0.
when you don't know or have no clue where is each of the two, you CANNOT use it.

For example in your graph i get that the mean value is positive, but i cannot be sure because it is not so obvious.

I saw no reason to post this thing ! Anyways in my original graph which was drawn in exam paper I could easily see that +x is many times greater than magnitude of x in -x. So mean value will be positive.