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Different parabola equation when finding area [Solved!]

My question

In chapter 3, compare example 1 to example 3. Why isnt (3 - x^2) the correct parabolic equation for example 3?

Relevant page

3. Areas Under Curves

What I've done so far

When I went to solve example 3, I used the equation `(3 - x^2)` for the parabola. I found out that was not the correct parabolic equation upon looking at your solution.
I am not sure why my equation cannot be used even though the vertex is not at (0,3), i.e. not lined up with the origin? The height is still three so I thought it would be okay.

X

In chapter 3, compare example 1 to example 3.  Why isnt (3 - x^2) the correct parabolic equation for example 3?
Relevant page

<a href="https://www.intmath.com/integration/3-area-under-curve.php">3. Areas Under Curves</a>

What I've done so far

When I went to solve example 3, I used the equation `(3 - x^2)` for the parabola. I found out that was not the correct parabolic equation upon looking at your solution.
I am not sure why my equation cannot be used even though the vertex is not at (0,3), i.e. not lined up with the origin?  The height is still three so I thought it would be okay.

Re: Different parabola equation when finding area

@phinah: Your parabola will certainly have the correct height, but let's consider its width. Where does your parabola intersect with the x-axis?

What width at the x-axis does this give you?

X

@phinah: Your parabola will certainly have the correct height, but let's consider its width. Where does your parabola intersect with the <i>x</i>-axis?

What width at the <i>x</i>-axis does this give you?

Re: Different parabola equation when finding area

I realized after I inquired to you that this upside down parabola is a shift one unit to the right on the x-axis.

So I came up with the `-(x-1)^2 + 3.`

But the parabola bottom would start at `(0,2)` Would that be an issue?

NOTE: The integral would be `(-x^2 + 2x + 2) dx` for a value of `16/3` square units.

X

I realized after I inquired to you that this upside down parabola is a shift one unit to the right on the x-axis.  

So I came up with the `-(x-1)^2 + 3.`  

But the parabola bottom would start at `(0,2)` Would that be an issue?

NOTE: The integral would be `(-x^2 + 2x + 2) dx` for a value of `16/3` square units.

Re: Different parabola equation when finding area

@phinah: I edited your answer to include math formatting, which you are encouraged to do.

Once again, the width at the `x`-axis is a problem (both with your original attempt and with this revision.)

Please try to answer my original questions:

a. Where does your parabola intersect with the x-axis when we use `y=3-x^2?`

b. What width at the `x`-axis does this give you?

You may find some of the background in this article useful: How to find the equation of a quadratic function from its graph

X

@phinah: I edited your answer to include math formatting, which you are encouraged to do.

Once again, the width at the `x`-axis is a problem (both with your original attempt and with this revision.)

Please try to answer my original questions: 

a. Where does your parabola intersect with the x-axis when we use `y=3-x^2?`

b. What width at the `x`-axis does this give you?

You may find some of the background in this article useful: <a href="https://www.intmath.com/blog/mathematics/how-to-find-the-equation-of-a-quadratic-function-from-its-graph-6070">How to find the equation of a quadratic function from its graph</a>

Re: Different parabola equation when finding area

`3 - x^2 = 0 `
` x = +/- sqrt 3'
So my length is 3.46 when it should only be 2.

X

`3 - x^2 = 0 `
` x = +/- sqrt 3'
So my length is 3.46 when it should only be 2.

Re: Different parabola equation when finding area

Correction

`x = + or - sqrt 3

X

Correction 

`x = + or - sqrt 3

Re: Different parabola equation when finding area

@Phinah: Yes, correct, so it's not possible to use `y=3-x^2` for your function. But it is possible to use one that passes through `(0, 3).` Can you find it?

BTW, to achieve "plus or minus" in the math entry system, you enter

+-

So your answer would look like: `x = +- sqrt(3)`.

You can see all the syntax for mathematical symbols here: ASCIIMath syntax.

X

@Phinah: Yes, correct, so it's not possible to use `y=3-x^2` for your function. But it is possible to use one that passes through `(0, 3).` Can you find it?

BTW, to achieve "plus or minus" in the math entry system, you enter

+-

So your answer would look like: `x = +- sqrt(3)`.

You can see all the syntax for mathematical symbols here: <a href="https://www.intmath.com/help/send-math-email-syntax.php">ASCIIMath syntax</a>.

Re: Different parabola equation when finding area

I cannot.

X

I cannot.

Re: Different parabola equation when finding area

Well, since we know the width of the arch must be 2 m, then we know it must go through `(-1, 0)` and `(1, 0).`

So using the general form of a quadratic,

`y = ax^2 + bx + c,`

you can substitute the 3 known points in and thus find the values of `a, b` and `c.`

Have a go!

X

Well, since we know the width of the arch must be 2 m, then we know it must go through `(-1, 0)` and `(1, 0).`

So using the general form of a quadratic,

`y = ax^2 + bx + c,`

you can substitute the 3 known points in and thus find the values of `a, b` and `c.`

Have a go!

Re: Different parabola equation when finding area

@murray: phinah hasn't responded. Would you like me to answer this?

X

@murray: phinah hasn't responded. Would you like me to answer this?

Re: Different parabola equation when finding area

@stephenB: Sure, please go ahead!

X

@stephenB: Sure, please go ahead!

Re: Different parabola equation when finding area

Okay, so we have the general formula

`y = ax^2 + bx + c,`

and the 3 points it needs to go through, `(-1, 0),` `(1, 0)` and `(0, 3).`

Plugging them in:

`0 = a(1)^2 + b(1) + c ` `= a + b + c ... [1]`

`0 = a(-1)^2 + b(-1) + c ` `= a - b + c ... [2]`

`3 = a(0)^2 + b(0) + c = c ... [3]`

Adding [1] and [2] gives

`0 = 2a + 2c` which means `a + c = 0 ...[4]`

From [3] we know `c=3`, so from [4], `a=-3` and then plugging those into [1] gives `b=0.`

That is, the parabola's equation is:

`y = -3x^2 + 3`

X

Okay, so we have the general formula

`y = ax^2 + bx + c,`

and the 3 points it needs to go through,  `(-1, 0),`  `(1, 0)` and `(0, 3).`

Plugging them in:

`0 = a(1)^2 + b(1) + c ` `= a + b + c ... [1]`

`0 = a(-1)^2 + b(-1) + c ` `= a - b + c ... [2]`

`3 = a(0)^2 + b(0) + c = c ... [3]`

Adding [1] and [2] gives

`0 = 2a + 2c` which means `a + c = 0 ...[4]`

From [3] we know `c=3`, so from [4], `a=-3` and then plugging those into [1] gives `b=0.`

That is, the parabola's equation is:

`y = -3x^2 + 3`

Re: Different parabola equation when finding area

Thanks, stephenB.

I think if phinah tries the equation you found, he will find the area of the parabolic glass pane will the the same as in my solution.

X

Thanks, stephenB.

I think if phinah tries the equation you found, he will find the area of the parabolic glass pane will the the same as in my solution.

Re: Different parabola equation when finding area

Phinah DID respond with the correct solution Murray BEFORE Stephen did. I do not know why it did not come through

X

Phinah DID respond with the correct solution Murray BEFORE Stephen did.  I do not know why it did not come through

Re: Different parabola equation when finding area

Hi Phinah. Sorry - it must have been a temporary database glitch or something.

X

Hi Phinah. Sorry - it must have been a temporary database glitch or something.

Re: Different parabola equation when finding area

Okay Murray. Thank you.

X

Okay Murray.  Thank you.

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