Using 3 Points to Find Quadratic Equation in General Form

[xpeteyzx]

Help! Using 3 Points to Find Quadratic Equation in General Form

I have a quadratic equation question that I have to solve, but I can't seem to understand it.
The question is: "Using points (1,0);(3,0);(0,-6), find the quadratic formula in General Form
(ax^2 + bx + c).

My teacher says that I have use the method of y= a(x-p) + q to solve it, no other method is allowed for me to use.

I had a try at it:

{ a(1-p)^2 + q= 0 ...(1)
{ a(3-p)^2 + q= 0 ...(2)
{ a(p)^2 + q= -6 ...(3)

{ a - 2ap + ap^2 + q= 0 ...(4)
{ 9a - 6ap + ap^2 + q= 0...(5)

{ -8a + 4ap= 0 ...(6)
{ -9a + 6ap= -6 ...(7)

{ -48a + 24ap= 0 ...(6)
{ -36a + 24ap= -6 ...(7)

{ -12a= 24 ...(8)
a= -2


I have the constant "a" down, but I can't get "p" or "q"
Any help is appreciated. Thanks a bunch!

 

[ehild]

Subtract equation 1 from eq 2.

ehild

 

[xpeteyzx]

I did that to get equation 6

 

[ehild]

You know a already, substitute it into eq,.6.ehild

 

[xpeteyzx]

Would you get P = -2?

 

[LCKurtz]

I have a quadratic equation question that I have to solve, but I can't seem to understand it.
The question is: "Using points (1,0);(3,0);(0,-6), find the quadratic formula in General Form
(ax^2 + bx + c).

My teacher says that I have use the method of y= a(x-p) + q to solve it, no other method is allowed for me to use.

You mean ##y = a(x-p)^2+q##

I had a try at it:
{ a(1-p)^2 + q= 0 ...(1)
{ a(3-p)^2 + q= 0 ...(2)
{ a(p)^2 + q= -6 ...(3)

{ a - 2ap + ap^2 + q= 0 ...(4)
{ 9a - 6ap + ap^2 + q= 0...(5)

{ -8a + 4ap= 0 ...(6)
{ -9a + 6ap= -6 ...(7)

{ -48a + 24ap= 0 ...(6)
{ -36a + 24ap= -6 ...(7)

{ -12a= 24 ...(8)
a= -2


I have the constant "a" down, but I can't get "p" or "q"

So put ##a=-2## in your equations 1 and 2 and see if you can solve them. It's easy.

 

[xpeteyzx]

ok, I'll try that

 

[ehild]

Why minus?

ehild

 

[e^(i Pi)+1=0]

(1,0);(3,0);(0,-6)

You must have made a typo; those points can't form a parabola.

 

[D H]

(1,0);(3,0);(0,-6)

You must have made a typo; those points can't form a parabola.

Yes, they do.

 

[symbolipoint]

What do you mean, "no other method is allowed"? Starting and proceeding through standard form is much more complicated. Why not use general form, create your three equations, perform row operations to find a, b, and c (as for general form), and then you can write the general form equation; from that, you can complete the square to find the standard form version of the parabola.

You would start with this matrix:
[ 1 1 1 0 ]
[ 9 3 1 0 ]
[ 0 0 1 -6 ]
For columns representing a, b, c, and y; the variables now being a, b, and c.

You would find the equation, y=-2x²+8x-6, and the three given points each satisfy this equation. (That is still in general form...)

 

[D H]

What do you mean, "no other method is allowed"?

He means he can't use the Vandermonde matrix approach that you outlined.

 

[Mentallic]

(1,0);(3,0);(0,-6)

You must have made a typo; those points can't form a parabola.

Given any 3 points, as long as they're not collinear and they all have different x-values to each other, you can form a parabola passing through them.

 

[Ray Vickson]

What do you mean, "no other method is allowed"? Starting and proceeding through standard form is much more complicated. Why not use general form, create your three equations, perform row operations to find a, b, and c (as for general form), and then you can write the general form equation; from that, you can complete the square to find the standard form version of the parabola.

You would start with this matrix:
[ 1 1 1 0 ]
[ 9 3 1 0 ]
[ 0 0 1 -6 ]
For columns representing a, b, c, and y; the variables now being a, b, and c.

You would find the equation, y=-2x²+8x-6, and the three given points each satisfy this equation. (That is still in general form...)

A much, much easier way is to note that the two zeros of the quadratic form are given as part of the input data, so y = A*(x-1)*(x-3). Get the constant A from the requirement that y = -6 when x = 0.

OK, that is not in the teacher's standard form, but it can easily be re-written to conform.

 

[skiller]

Given any 3 points, as long as they're not collinear and they all have different x-values to each other, you can form a parabola passing through them.

Not strictly true. The curve [itex]x=y^2[/itex] is also a parabola, isn't it?

 

[D H]

Not strictly true. The curve [itex]x=y^2[/itex] is also a parabola, isn't it?

Strictly true.

If you generate three (x,y) pairs from x=y² such that all of the x values are different you won't recreate x=y² when trying to fit to y=ax^2+bx+c. You will however find a unique set of values for a, b, and c.

 

[Mentallic]

Not strictly true. The curve [itex]x=y^2[/itex] is also a parabola, isn't it?

Yes Although I left it out to avoid going on and on about the situations of when it's a function of x versus a function of y, and we could have even gone further and mentioned rotated parabolas etc.

 

[symbolipoint]

A much, much easier way is to note that the two zeros of the quadratic form are given as part of the input data, so y = A*(x-1)*(x-3). Get the constant A from the requirement that y = -6 when x = 0.

OK, that is not in the teacher's standard form, but it can easily be re-written to conform.

Yes, I almost had that but become confused during that analysis. Although I solved the problem in the way as I indicated earlier, not the way xpetey wanted, at least converting into standard form made sense.

 

[skiller]

Strictly true.

If you generate three (x,y) pairs from x=y² such that all of the x values are different you won't recreate x=y² when trying to fit to y=ax^2+bx+c. You will however find a unique set of values for a, b, and c.

Ok, strictly true. The 3 x-values being different is a sufficient condition but it isn't a necessary condition, which is what I thought Mentallic was suggesting (and his/her subsequent response supports that).

But I've no idea why you're trying to fit any general parabola to [itex]y=ax^2+bx+c[/itex]

That may be relevant to this thread's OP, but not what I was saying to Mentallic.

 

[PeterO]

I have a quadratic equation question that I have to solve, but I can't seem to understand it.
The question is: "Using points (1,0);(3,0);(0,-6), find the quadratic formula in General Form
(ax^2 + bx + c).

My teacher says that I have use the method of y= a(x-p) + q to solve it, no other method is allowed for me to use.

I had a try at it:

{ a(1-p)^2 + q= 0 ...(1)
{ a(3-p)^2 + q= 0 ...(2)
{ a(p)^2 + q= -6 ...(3)

{ a - 2ap + ap^2 + q= 0 ...(4)
{ 9a - 6ap + ap^2 + q= 0...(5)

{ -8a + 4ap= 0 ...(6)
{ -9a + 6ap= -6 ...(7)

{ -48a + 24ap= 0 ...(6)
{ -36a + 24ap= -6 ...(7)

{ -12a= 24 ...(8)
a= -2


I have the constant "a" down, but I can't get "p" or "q"
Any help is appreciated. Thanks a bunch!

Was your teacher requiring you to use that form in solution, or suggesting that would be the best way.

Those 2 points (1,0) and (3,0) make that method pretty useful,
but then the form y = a(x-m)(x-n) is also pretty handy.

 

[jtbell]

Duplicate threads merged.

 

[xpeteyzx]

I am getting really confused...

 

[xpeteyzx]

Was your teacher requiring you to use that form in solution, or suggesting that would be the best way.

Those 2 points (1,0) and (3,0) make that method pretty useful,
but then the form y = a(x-m)(x-n) is also pretty handy.

He told me that I had to use only that theory to solve it. He said any other theory results in a F for that assignment.

 

[xpeteyzx]

Was your teacher requiring you to use that form in solution, or suggesting that would be the best way.

Those 2 points (1,0) and (3,0) make that method pretty useful,
but then the form y = a(x-m)(x-n) is also pretty handy.

The equation theory of y=a(x-p) + q

 

[D H]

The equation theory of y=a(x-p) + q

You've been asked this before, but you never did answer: Are you *sure* your teacher didn't tell you to use y=a(x-p)²+q instead of y=a(x-p)+q ?

 

[symbolipoint]

You've been asked this before, but you never did answer: Are you *sure* your teacher didn't tell you to use y=a(x-p)²+q instead of y=a(x-p)+q ?

xpeteyzx,

That y=a(x-p)²+q is just the standard form for a parabola. It permits more easily translating between graph and equation. Using it directly may more easily allow you to find value of "a", but you can still use general form to setup your three equations and solve as a matrix for b and c of the GENERAL form; and then convert to standard form.

 

[D H]

but you can still use general form to setup your three equations and solve as a matrix for b and c of the GENERAL form; and then convert to standard form.

My take is that this approach will get him an F. As I understand the problem, he needs to determine the values of a, p, and q in a(x-p)²+q and then convert those to the a, b, and c in y=ax²+bx+c.

 

[symbolipoint]

I am imagining a way, starting from purely factored form as Ray Vickson said. Completing the square on axY2+bx+c and comparing the two standard forms will show p and q according to corresponding positions of variables.

 

[D H]

I think that we should focus on teaching how to find the values of a, p, and q in y=a(x-p)²+q given that y(1)=0, y(3)=0, and y(0)=-6.

xpeteyzx, try using the fact that y(1)=y(3) to find p. Then use y(0)=-6 and either of y(1)=0 or y(3)=0 to find the values of a and q.

 

[xpeteyzx]

You've been asked this before, but you never did answer: Are you *sure* your teacher didn't tell you to use y=a(x-p)²+q instead of y=a(x-p)+q ?

Sorry, I meant y=a(x-p)^2 + q

 

[LCKurtz]

I think that we should focus on teaching how to find the values of a, p, and q in y=a(x-p)²+q given that y(1)=0, y(3)=0, and y(0)=-6.

Wasn't that done long before this/these thread(s) want bananas? Post #6, which I think was post #2 or #3 before the threads were merged did just that.