Math Is Fun Forum

mathland

Member

Registered: 2021-03-25

Posts: 444

Find a & b

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.)

1. How is this done?
2. If there are many correct answers, what's the objective here?
3. What are the rules of symmetry with respect to the x-axis, y-axis
and the origin?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,580

Re: Find a & b

hi mathland,

I think it will make more sense if I answer your questions in reverse order.

Symmetry with respect to a line (such as one of the axes) means that line is a line of symmetry.  Sometimes this is described as a mirror line.

Symmetry with respect to the origin means there is rotational symmetry order 2.  If the whole graph is rotated 180 degrees around the origin the graph fits over itself.

If x = 0 in the equation then y = 0 whatever the values of a and b, so that's a good start for all three types of symmetry.

A quadratic in x  always has a vertical line of symmetry.  [You might want to consider why this is true.]

Cubic equations ( highest power of x is 3) will always start in the third quadrant and end in the first (when the cube term is positive) or start in the second and end in the fourth (when the cube term is negative)

Why ask a question which has many answers?

I think the question is designed to make you think about the above, testing your understanding of the types of symmetry, and the  properties of quadratics and cubics in general.

So finally the first part:

A cubic cannot have a line of symmetry so think how you can 'remove' the cubic term to get the y axis as a line of symmetry.

To test for rotational symmetry around the origin you need f(-x) = -f(x) so that if (p,q) lies on the graph then so does (-p,-q).

There is one and only one value for a that will allow this to happen. If you cannot already see it try plotting the graph with a variety of values for a and b.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Find a & b

hi mathland,

I think it will make more sense if I answer your questions in reverse order.

Symmetry with respect to a line (such as one of the axes) means that line is a line of symmetry.  Sometimes this is described as a mirror line.

Symmetry with respect to the origin means there is rotational symmetry order 2.  If the whole graph is rotated 180 degrees around the origin the graph fits over itself.

If x = 0 in the equation then y = 0 whatever the values of a and b, so that's a good start for all three types of symmetry.

A quadratic in x  always has a vertical line of symmetry.  [You might want to consider why this is true.]

Cubic equations ( highest power of x is 3) will always start in the third quadrant and end in the first (when the cube term is positive) or start in the second and end in the fourth (when the cube term is negative)

Why ask a question which has many answers?

I think the question is designed to make you think about the above, testing your understanding of the types of symmetry, and the  properties of quadratics and cubics in general.

So finally the first part:

A cubic cannot have a line of symmetry so think how you can 'remove' the cubic term to get the y axis as a line of symmetry.

To test for rotational symmetry around the origin you need f(-x) = -f(x) so that if (p,q) lies on the graph then so does (-p,-q).

There is one and only one value for a that will allow this to happen. If you cannot already see it try plotting the graph with a variety of values for a and b.

Bob

To remove the cubic term, I must divide all terms by x^2. Yes?

You said:

"There is one and only one value for a that will allow this to happen. If you cannot already see it try plotting the graph with a variety of values for a and b."

Can you explain this statement?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,580

Re: Find a & b

To remove the cubic term, I must divide all terms by x^2. Yes?

Sorry, but no.  That introduces a term y/(x^2) so doesn't help.  I intended that you do it by choosing a suitable value for 'b'.

To make the function have the property if (p,q) lies on the curve then so does (-p,-q)  proceed as follows:

         and

It's not too hard to eliminate both q and b by adding these two equations.  That gives the required value for a.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Find a & b

To remove the cubic term, I must divide all terms by x^2. Yes?

Sorry, but no.  That introduces a term y/(x^2) so doesn't help.  I intended that you do it by choosing a suitable value for 'b'.

To make the function have the property if (p,q) lies on the curve then so does (-p,-q)  proceed as follows:

         and

It's not too hard to eliminate both q and b by adding these two equations.  That gives the required value for a.

Bob

Hi Bob. What's new with you? I apologize for not making it clear in terms of where the posted questions come from. I can easily answer chapter questions for which the author(s) has or have given samples for.

The questions posted here come from the CHALLENGE SECTION. According to Michael Sullivan, if the particular section has been fully understood, then students should be able to answer applied problems requiring more thinking. I find that this isn't always the case. You say?