1. ## Parabola Problem

A parabola has x-axis values of -3 and 5. And it symmetry axis is: x = 1

The a value is negative, but i cannot say what its exact value is.

It says to prove that $f(x) = -x^2 + 2x + 15$

Here's what i did:

$f(x) = a(x + 3)(x - 5)$

$f(x) = a(x^2 - 2x - 15)$

And now i KNOW that $a$ must be equal to $-1$ but i have no idea how to prove it

EDIT: Sorry, i forgot to mention that there is also a graph $g(x) = |x| - 3$

2. Originally Posted by janvdl
A parabola has x-axis values of -3 and 5. And it symmetry axis is: x = 1

The a value is negative, but i cannot say what its exact value is.

It says to prove that $f(x) = -x^2 + 2x + 15$

Here's what i did:

$f(x) = a(x + 3)(x - 5)$

$f(x) = a(x^2 - 2x - 15)$

And now i KNOW that $a$ must be equal to $-1$ but i have no idea how to prove it

EDIT: Sorry, i forgot to mention that there is also a graph $g(x) = |x| - 3$
I presume that "x-axis values of -3 and 5" means these are the x intercepts (-3, 0) and (5, 0)?

Then
$f(x) = a(x^2 - 2x - 15)$
is as good as you are going to get with the current information. The only thing you can do with it now is to find the axis of symmetry, which is supposed to be the line x = 1. But:
$f(x) = a(x^2 - 2x - 15) = ax^2 + -2ax - 15a$
has an axis of symmtry at
$x = -\frac{-2a}{2 \cdot a} = 1$
and the "a" cancels out naturally. So we have no new information here.

What is the significance of the graph of $g(x) = |x| - 3$?

-Dan

3. Originally Posted by topsquark
I presume that "x-axis values of -3 and 5" means these are the x intercepts (-3, 0) and (5, 0)?

Then
$f(x) = a(x^2 - 2x - 15)$
is as good as you are going to get with the current information. The only thing you can do with it now is to find the axis of symmetry, which is supposed to be the line x = 1. But:
$f(x) = a(x^2 - 2x - 15) = ax^2 + -2ax - 15a$
has an axis of symmtry at
$x = -\frac{-2a}{2 \cdot a} = 1$
and the "a" cancels out naturally. So we have no new information here.

What is the significance of the graph of $g(x) = |x| - 3$?

-Dan
Apparently i'm supposed to use $f(x) = a(x - p)^2 + q$

I'm still busy figuring it out, i'm quite tired

Thanks anyway Topsquark.

4. Originally Posted by janvdl
Apparently i'm supposed to use $f(x) = a(x - p)^2 + q$

I'm still busy figuring it out, i'm quite tired

Thanks anyway Topsquark.
That form would be helpful if you
1) Had three points on the parabola
or
2) Knew the vertex of the parabola

Since neither of these conditions are met, you can't get any further.

-Dan

6. The problem is that
$f(x) = a(x^2 - 2x - 15)$

$f(x) = a(x^2 - 2x + 1) - a(1) - 15a$

$f(x) = a(x - 1)^2 - 16a$
has multiple solutions in terms of what you are being asked for. Any value of "a" will give you a parabola with the given properties.

-Dan

7. Originally Posted by topsquark
The problem is that
$f(x) = a(x^2 - 2x - 15)$

$f(x) = a(x^2 - 2x + 1) - a(1) - 15a$

$f(x) = a(x - 1)^2 - 16a$
has multiple solutions in terms of what you are being asked for. Any value of "a" will give you a parabola with the given properties.

-Dan
So the memo is wrong?

8. Originally Posted by janvdl
So the memo is wrong?
I couldn't read it that well, but it appeared to give a = -1 explicitly and I can construct the parabola for a = -1/2, a = -1/sqrt(3), etc. with intercepts at x = -3 and 5 with an axis of symmetry x = 1 (as well as for a = -1.)

-Dan

9. Originally Posted by topsquark
I couldn't read it that well, but it appeared to give a = -1 explicitly and I can construct the parabola for a = -1/2, a = -1/sqrt(3), etc. with intercepts at x = -3 and 5 with an axis of symmetry x = 1 (as well as for a = -1.)

-Dan
But then a shouldn't necessarily only be -1 ?

Then a could be any value? So it could also have an equation of $f(x) = -2x^2 ... \ -3x^2 ... \ -4x^2... ?$

This is the weirdest parabola prob in all my high school years...

10. Originally Posted by janvdl
But then a shouldn't necessarily only be -1 ?

Then a could be any value? So it could also have an equation of $f(x) = -2x^2 ... \ -3x^2 ... \ -4x^2... ?$

This is the weirdest parabola prob in all my high school years...
Yup. Not only that, "a" could also be positive and give a "correct" parabola. The only reason it can't be is that problem told you to assume it wasn't.

It's only this weird because they didn't specify enough information. If another condition on the parabola were given then we could set up an equation to solve for "a."

-Dan