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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.)

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.)

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

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]

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]

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.