Find a and b to make this equation symmetric about the y-axis: y = ax^2 + bx^3

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  • #1
nycmathguy
Homework Statement:
Math HW help for a friend.
Relevant Equations:
Symmetry with respects to the y-axis and origin.
My friend asked for help with this precalculus question. I could not help him. So, I decided to post here.

Find a and b when the graph of
y = ax^2 + bx^3 is symmetric with respect to (a) the y-axis and (b) the origin. (There are many correct answers.)

I don't even know where to begin.
 

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  • #2
Delta2
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Do you know what conditions should hold so that the graph has these symmetries? They are some equalities that relate y(x) and y(-x).
 
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  • #3
nycmathguy
Do you know what conditions should hold so that the graph has these symmetries? They are some equalities that relate y(x) and y(-x).

Are you saying to replace x with -x?
What about replacing y with -y?
Can you set it up for me?
 
  • #4
Delta2
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well something like that.

In order to be more accurate, the condition for symmetry around the y-axis is
y(x)=y(-x) for all x.
In order to have symmetry around the origin the condition is
y(x)=-y(-x) for all x.

Find out what are the allowed values for a and b such that these conditions hold.
 
  • #5
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Are you saying to replace x with -x?
Let ##f(x) = ax^2 + bx^3##
If ##f(x) = f(-x)##, the graph is symmetric about the y-axis. In other word, the points (c, f(c)) and (-c, f(-c)) are on the graph.
If ##f(x) = -f(-x)##, the graph is symmetric about the origin. IOW, the points (c, f(c)) and (-c, -f(-c)) are on the graph.

If a graph is symmetric about the x-axis, it's not a function.

As I see it, this is a very simple problem. There are infinitely many possible solutions, but there is a pair of very simple solutions.

Quadratic functions are inherently symmetric about the y-axis unless they have been translated. Cubic functions are inherently symmetric about the origin, also unless they have been translated.
 
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  • #6
nycmathguy
well something like that.

In order to be more accurate, the condition for symmetry around the y-axis is
y(x)=y(-x) for all x.
In order to have symmetry around the origin the condition is
y(x)=-y(-x) for all x.

Find out what are the allowed values for a and b such that these conditions hold.

The question goes on to say there are different acceptable answers.
 
  • #7
nycmathguy
Let ##f(x) = ax^2 + bx^3##
If ##f(x) = f(-x)##, the graph is symmetric about the y-axis. In other word, the points (c, f(c)) and (-c, f(-c)) are on the graph.
If ##f(x) = -f(-x)##, the graph is symmetric about the origin. IOW, the points (c, f(c)) and (-c, -f(-c)) are on the graph.

If a graph is symmetric about the x-axis, it's not a function.

As I see it, this is a very simple problem. There are infinitely many possible solutions, but there is a pair of very simple solutions.

Quadratic functions are inherently symmetric about the y-axis unless they have been translated. Cubic functions are inherently symmetric about the origin, also unless they have been translated.

You said:

"If a graph is symmetric about the x-axis, it's not a function."

Why is this the case?

You said:

"As I see it, this is a very simple problem. There are infinitely many possible solutions, but there is a pair of very simple solutions."

Sir, this is true if and only if you are gifted mathematically or math is or was your major. Trust me, for most students, this question is not so clear. According to my friend, this question is in the challenging section in his Ron Larson Precalculus Edition 10E textbook.

I play hymns on the guitar in the classical style. I started playing guitar in 1977. I was 12 years old. I also read and write guitar tablature. If you wanted to learn how to play in the classical guitar style, I would never say COME ON, IT'S EASY. ANYBODY CAN DO WHAT I DO ON THE GUITAR. Understand? If you know something, anything it's easy.
 
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  • #8
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You said:

"If a graph is symmetric about the x-axis, it's not a function."

Why is this the case?
Because of how functions are defined. I.e., each number in the domain of the function is paired with exactly one number in the range. The graph of the equation ##y^2 = x## is symmetric about the x-axis, so this equation represents a relation but not a function.
nycmathguy said:
You said:
"As I see it, this is a very simple problem. There are infinitely many possible solutions, but there is a pair of very simple solutions."

Sir, this is true if and only if you are gifted mathematically or math is or was your major. Trust me, for most students, this question is not so clear. According to my friend, this question is in the challenging section in his Ron Larson Precalculus Edition 10E textbook.
For people whose goal is studying calculus, like yourself, there needs to be a solid grounding in precalculus topics such as the graphs of simple functions and the kinds of symmetry that various functions have. One should also understand the importance of definitions, such as how functions are defined and how the various kinds of symmetry are defined.
 
  • #9
Gaussian97
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If ##f(x) = f(-x)##, the graph is symmetric about the y-axis. In other word, the points (c, f(c)) and (-c, f(-c)) are on the graph.
If ##f(x) = -f(-x)##, the graph is symmetric about the origin. IOW, the points (c, f(c)) and (-c, -f(-c)) are on the graph.
Aren't the points ##(c,f(c))## and ##(-c,f(-c))## in the graph of any function?
Also, for ##\sin(x)##, the point ##\left(-\frac{\pi}{2}, -\sin\left(-\frac{\pi}{2}\right)\right)## is not on the graph, isn't ##\sin(x)## symmetric?
 
  • #10
nycmathguy
Because of how functions are defined. I.e., each number in the domain of the function is paired with exactly one number in the range. The graph of the equation ##y^2 = x## is symmetric about the x-axis, so this equation represents a relation but not a function.
For people whose goal is studying calculus, like yourself, there needs to be a solid grounding in precalculus topics such as the graphs of simple functions and the kinds of symmetry that various functions have. One should also understand the importance of definitions, such as how functions are defined and how the various kinds of symmetry are defined.

Read carefully.

1. This is my friend's math HW question not mine. I am 56. He is 28. Ok?

2. I took Precalculus in the Spring 1993 semester at Lehman College in the Bronx NY. I got an A minus in the course. It was not a required course for graduation. I majored in Sociology not Mathematics.

3. I took precalculus as an elective course then. I have no idea if Lehman still has elective courses. Anyway, 2021 - 1993 = how long it's been since I last "played" with material pertaining to this course. I was not able to help my friend with this particular question but I still remember precalculus to a certain extent.

4. So, your questions should be: WHY STUDY CALCULUS 1 NOW? WHY NOT REVISIT PRECALCULUS? I am 56. I have no idea how long I have left to live. Revisiting precalculus would take me forever to complete a textbook.

5. I still recall most of my precalculus notes but I could not help my friend find a and b for THIS PARTICULAR question.
 
  • #11
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Aren't the points ##(c,f(c))## and ##(-c,f(-c))## in the graph of any function?
Yes, of course, but for a function that is symmetric about the y-axis, f(c) = f(-c), which I wrote in the immediately preceding sentence.
Also, for ##\sin(x)##, the point ##\left(-\frac{\pi}{2}, -\sin\left(-\frac{\pi}{2}\right)\right)## is not on the graph, isn't ##\sin(x)## symmetric?
You're right. I wrote the definition correctly, but added an extra minus sign that shouldn't be there in my IOW part.
 
  • #12
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1. This is my friend's math HW question not mine. I am 56. He is 28. Ok?
Why does this matter?

2. I took Precalculus in the Spring 1993 semester at Lehman College in the Bronx NY. I got an A minus in the course. It was not a required course for graduation. I majored in Sociology not Mathematics.

3. I took precalculus as an elective course then. I have no idea if Lehman still has elective courses. Anyway, 2021 - 1993 = how long it's been since I last "played" with material pertaining to this course. I was not able to help my friend with this particular question but I still remember precalculus to a certain extent.
That's 28 years. Since it has been quite a while, it would be good to review the precalc topics.

4. So, your questions should be: WHY STUDY CALCULUS 1 NOW? WHY NOT REVISIT PRECALCULUS? I am 56. I have no idea how long I have left to live. Revisiting precalculus would take me forever to complete a textbook.
If you are rusty on the precalc topics it will be impossible to make any progress in calculus.

5. I still recall most of my precalculus notes but I could not help my friend find a and b for THIS PARTICULAR question.
What does the graph of ##y = x^2## look like? What does the graph of ##y = x^3## look like? These are some of the basic ideas I was talking about earlier.
 
  • #13
nycmathguy
Why does this matter?

That's 28 years. Since it has been quite a while, it would be good to review the precalc topics.

If you are rusty on the precalc topics it will be impossible to make any progress in calculus.

What does the graph of ##y = x^2## look like? What does the graph of ##y = x^3## look like? These are some of the basic ideas I was talking about earlier.
Ok. I will do the following:

1. I will review precalculus while at the Same time learning Calculus l at a slower paste.

2. See below for graphs.
 

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  • #14
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1. I will review precalculus while at the Same time learning Calculus l at a slower paste.
That's a good plan. You also don't need to go through the precalc text page by page. Look at the review/summary questions at the ends of the chapters, and see if you can work some of the problems. Many of them will have answers in the back so you can check your work.

2. See below for graphs.
Is either of them symmetric about the y-axis? About the origin?
 
  • #15
nycmathguy
That's a good plan. You also don't need to go through the precalc text page by page. Look at the review/summary questions at the ends of the chapters, and see if you can work some of the problems. Many of them will have answers in the back so you can check your work.

Is either of them symmetric about the y-axis? About the origin?


Algebraic Tests for Symmetry

1. The graph of an equation is symmetric with respect to the x-axis when replacing y with −y yields an equivalent equation.

2. The graph of an equation is symmetric with respect to the y-axis when replacing x with −x yields an equivalent equation.

3. The graph of an equation is symmetric with respect to the origin when replacing x with −x and y with −y yields an equivalent equation.

For y = x^2

Let x = -x

y = (-x)^2

y = x^2

Symmetric about the y-axis aka the line x = 0.

Let y = -y

-y = x^2

y = -x^2

Not symmetric about the x-axis aka the y = 0 line.

Let x = -x and y = -y

-y = (-x)^2

-y = x^2

Not symmetric about the origin aka (0, 0).

You say?

For y = x^3

Let y = -y

-y = x^3

y = -x^3

Not symmetric about the x-axis.

Let x = -x

y = (-x)^3

y = -x^3

Not symnetric about the y-axis.

Let x = -x and y = -y

-y = (-x)^3

-y = -x^3

y = (-x^3)/(-1)

y = x^3

Symmetric about the origin.

You say?

I will post more precalculus questions later.
I don't mind going through an entire precalculus textbook. I honestly think it will be beneficial. However, calculus 1 questions will also be posted. It will be like taking two courses just for fun. What else am I to do at 56?

A. Fat
B. Bald
C. Divorced
D. Single
E. Not popular with PRETTY females

Should I go on? Now, back to math.
 
  • #16
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3. The graph of an equation is symmetric with respect to the origin when replacing x with −x and y with −y yields an equivalent equation.

For y = x^2

Let x = -x
I get what you're doing, but it's incorrect/confusing to say "Let x = -x".
Better...
For ##f(x) = x^2##,
##f(-x) = (-x)^2 = x^2 = f(x)##
Therefore this function is symmetric about the y-axis.

For y = x^3

Let y = -y
Same problem...
Better...
For ##g(x) = x^3##
##g(-x) = (-x)^3 = -x^3 = -g(x)##
Therefore this function is symmetric about the origin.
 
  • #17
nycmathguy
I get what you're doing, but it's incorrect/confusing to say "Let x = -x".
Better...
For ##f(x) = x^2##,
##f(-x) = (-x)^2 = x^2 = f(x)##
Therefore this function is symmetric about the y-axis.

Same problem...
Better...
For ##g(x) = x^3##
##g(-x) = (-x)^3 = -x^3 = -g(x)##
Therefore this function is symmetric about the origin.
Yes, this is what I was trying to say. Now, back to the original question for my friend. What is the value of a and b? I still cannot figure it out. I posted below for your convenience.

Find a and b when the graph of
y = ax^2 + bx^3 is symmetric with respect to (a) the y-axis and (b) the origin. (There are many correct answers.)
 
  • #18
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Find a and b when the graph of
y = ax^2 + bx^3 is symmetric with respect to (a) the y-axis and (b) the origin. (There are many correct answers.)
What kind of graph do you get when a = 0?
 
  • #19
nycmathguy
What kind of graph do you get when a = 0?

When a = 0, we are left with y = bx^3, a cubic function.
 
  • #21
nycmathguy
What kind of graph do you get when a = 0?

Hello Mark. My friend Mira responded to my friend's question in my FB Precalculus group.

Here is Mira's reply. Tell me what you think.

Find a and b when the graph of
y = ax^2 + bx^3 is symmetric with respect to

(a) the y-axis

A graph is said to be symmetric about the y -axis if whenever (a,b) is on the graph then so is (−a,b). So, y = ax^2 + bx^3 implies (a,b).
Also, y = a(-x)^2 + b(-x)^3 = ax^2 - bx^3 implies
(-a,b).

Note: ax^2 + bx^3 = ax^2 - bx^3
ax^2 + bx^3-ax^2+ bx^3=0
bx^3+ bx^3=0
2bx^3=0=> always true if b=0.

Hence, y = ax^2 + 0*x^3
y = ax^2 =>the graph will be symmetric with respect to the y-axis.

(b) the origin.

The graph will be symmetric with respect to the origin if replacing x with -x, and y with -y not change equation -y = a(-x)^2 + b(-x)^3

So, -y = ax^2 - bx^3....multiply both sides by -1
y = -ax^2 + bx^3=>this must be equal to original equation.

We say,
ax^2 + bx^3= -ax^2 + bx^3
ax^2 + bx^3+ax^2 - bx^3=0
2ax^2=0 -> always true if a=0.

Since equation is independent of b, it can take any value.

Solution: a=0, b=>can take any value
There are many correct answers.
 
  • #22
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Pretty much what I had in mind, but much wordier.

Edited to fix errors
If a = 0 and b = 1, we have f(x) = x3, which is symmetric about the origin. It's easy to show that f(-x) = -f(x) for all x.
If a = 1 and b = 0, we have f(x) = x2, which is symmetric about the y-axis. It's easy to show that f(-x) = f(x) for all x.
 
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  • #23
nycmathguy
Pretty much what I had in mind, but much wordier.

Edited to fix errors
If a = 0 and b = 1, we have f(x) = x3, which is symmetric about the origin. It's easy to show that f(-x) = --f(x) for all x.
If a = 1 and b = 0, we have f(x) = x2, which is symmetric about the y-axis. It's easy to show that f(-x) = f(x) for all x.

Note: f(-x) = f(x) for all x ...any negative number squared will be positive. So, f(x)= f(-x) if f(x) = x^2.
 
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  • #24
haruspex
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Pretty much what I had in mind, but much wordier.

If a = 0 and b = 1, we have f(x) = x3, which is symmetric about the origin. It's easy to show that f(-x) = f(x) for all x.
If a = 1 and b = 0, we have f(x) = x2, which is symmetric about the y-axis. It's easy to show that f(-x) = -f(x) for all x.
Did you mean that the other way around, f(-x)=-f(x) for f(x) = x3 and f(-x) = f(x) for f(x) = x2?
 
  • #25
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Did you mean that the other way around, f(-x)=-f(x) for f(x) = x3 and f(-x) = f(x) for f(x) = x2?
Yes. I wrote the opposite of what I meant. I have edited my earlier reply.
 
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