#### judocallin

##### New member

- Joined
- Jan 18, 2009

- Messages
- 12

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.

- Thread starter judocallin
- Start date

- Joined
- Jan 18, 2009

- Messages
- 12

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.

- Joined
- Oct 6, 2005

- Messages
- 10,255

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

MY EDIT: Corrected name to General Form

- Joined
- Jan 18, 2009

- Messages
- 12

- Joined
- Oct 6, 2005

- Messages
- 10,255

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

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

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.)

- Joined
- Jan 18, 2009

- Messages
- 12

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

4A + 2B = 1

A + B = -2

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

Is that correct?

Is C=-4?

sorry for the very late response...

- Joined
- Jan 18, 2009

- Messages
- 12

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

Sorry, I can't remember his suggested answer.

- Joined
- Oct 6, 2005

- Messages
- 10,255

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

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

- Joined
- Jan 18, 2009

- Messages
- 12

-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

- Joined
- Oct 6, 2005

- Messages
- 10,255

judocallin said: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.

- Joined
- Jan 18, 2009

- Messages
- 12

funny!!! :lol: