View Full Version : finding the quadratic equation HELP!....thanks

judocallin

01-25-2010, 07:30 PM

Good Day! We are given this assingment,

Find the quadratic equation with the value of y=-3 at x=2, y=-6 at x=1 and y=-4 at x=0

unfortunately,I'm Stuck!

thanks for any help.

mmm4444bot

01-25-2010, 07:44 PM

Have you learned about solving systems of equations?

We can create a system of three equations, by substituting the given pairs of (x,y) values into one of the quadratic forms.

For example, we could use the General Form of a Quadratic Equation: y = Ax^2 + Bx + C

(0, -4) ? -4 = A(0)^2 + B(0) + C

(1, -6) ? -6 = A(1)^2 + B(1) + C

(2, -3) ? -3 = A(2)^2 + B(2) + C

This is a system of three equations to solve for A, B, and C.

If you need more help, please show us what you've learned about solving systems like this.

Cheers ~ Mark 8-)

MY EDIT: Corrected name to General Form

judocallin

01-26-2010, 01:49 AM

Say for example, I've already solved for the values of A,B,C. How can these values lead me to just one equation?Am I just going to substitute these values?If so, what value should I place for the value of y?Am I going to use the form ax^2 +bx+c=0, to solved this?thanks.=)

mmm4444bot

01-26-2010, 02:03 AM

Am I going to use the form ax^2 +bx+c=0, to solved this? No. There is nothing "to solve", in this manner.

This exercise is not like being told to find the values of x that satisfy something like 7x^2 - 10x - 3 = 0 (in other words, having to find x when y is 0). In this exercise, y is not zero.

All they want to see is the General Form of a quadratic equation [ y = Ax^2 + Bx + C ] with the correct Real values substituted for those three parameter symbols A, B, and C.

But, the Real values that you substitute for the parameters A, B, and C need to produce a correct formula for y. In other words, the quadratic equation that you write must be for the particular parabola that passes through those three given points.

You will know when the formula is correct because, when you use it to calculate the value of y, you will get -4 when x is 0, you will get -6 when x is 1, and you will get -3 when x is 2.

That's why I used the symbols A, B, and C as variables, in my first response, to generate those three equations, where y = -4 when x = 0, and y = -6 when x = 1, and y = -3 when x = 2.

If those three Real numbers for A, B, and C work in the system, then they will work in the quadratic equation to produce the given coordinates (0,-4) (1,-6) 2,-3).

Does it make more sense now?

Solve the system of three equations in my first response. Substitute the solution for A,B,C in the General Form, and write down that equation. You're done.

Do you know how to solve a system of two equations in two unknowns? Like the following.

4A + 2B = 1

A + B = -2

MY EDIT: Corrected name to General Form :roll: (I'm currently wondering who really cares what it's called.)

judocallin

02-07-2010, 07:01 AM

Do you know how to solve a system of two equations in two unknowns? Like the following.

4A + 2B = 1

A + B = -2

Yes!

The answer is A =5/2 B=-9/2

Is that correct?

Is C=-4?

sorry for the very late response...

Aladdin

02-07-2010, 07:21 AM

------------------

Yes !

judocallin

02-07-2010, 07:33 AM

But our professor's answer is different. He said it can be solve by a quadratic trick. What is that trick?

I tried searching for it but I can't find it.

Sorry, I can't remember his suggested answer.

mmm4444bot

02-08-2010, 01:39 AM

Those are the correct values of the parameters A, B, and C, for the quadratic polynomial equation (in general form) whose graph passes through the three points that you provided.

Does the answer provided by the instructor look like the following (standard or vertex form)?

y = \frac{5}{2} \left (x - \frac{9}{10} \right)^2 - \frac{241}{40}

judocallin

02-12-2010, 12:36 AM

my classmate said the trick is get the x coordinates and

-3 -6 -4 subtract -3 and -6

...\/ \/ subtract -6 and -4

.. 3 -2

....\/ subtract 3 and -2

....5

divide 5 by 2 to get "A"

to get "B" subtract -2 by "A"

and -4 is the last term

giving is the naswer (5/2)x^2-(9/2)x-4

mmm4444bot

02-12-2010, 01:28 AM

divide 5 by 2 to get "A"

to get "B" subtract -2 by "A"

and -4 is the last term

Good grief. :roll:

It works, in this case, but this trick does not work, in most cases.

giving is the naswer (5/2)x^2-(9/2)x-4 This is the same result that you earlier posted was wrong, according to your professor!

We already told you that it's correct, for the three points that you provided.

judocallin

02-12-2010, 02:05 AM

funny!!! :lol:

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