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.

@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?

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.

@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>

@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>.

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!

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`

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.

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