# Construct quadratic equation using the vertex method

1. Sep 7, 2012

### bugatti79

1. The problem statement, all variables and given/known data

Folks,
I wish to construct a quadratic expression from a graph which has the following 2 coordinates.

$(0,0)$ and the vertex $(\alpha h_e,1)$ where $0<\alpha<1$

2. Relevant equations

Using the following $y(x)=a(x-h)^2+k$

3. The attempt at a solution

$(h,k)=(\alpha h_e,1)$

From the first coordinate pair we have $\displaystyle 0=a(0-\alpha h_e )^2+1 \implies a=-\frac{1}{\alpha^2 h_e^2}$

but the coefficient of x^2 given in the book doesnt seem to tally up. Its given in the form

$\displaystyle \frac{x^2}{((-1 + a) a h^2)}-\frac{x}{((-1 + a) a h)}$

what have I done wrong...?

thanks

2. Sep 7, 2012

### Staff: Mentor

What is $h_{\epsilon}$?

What I don't see in your work is the idea that a parabola is symmetric across its central axis, which in this case would be the vertical line x = $\alpha h_{\epsilon}$

You are given that (0, 0) is one point on the parabola - there's another point directly across the vertical axis.

3. Sep 7, 2012

### SammyS

Staff Emeritus
The general expression for a quadratic function with vertex at (h, k) is
a(x - h)2 + k .​

In your case the h is $\alpha h_e$ and the k is 1 .

So you have $f(x)=a(x-\alpha h_e)^2 + 1\ .$

Since you know that the graph of y = f(x) passes through the point (0, 0), it must be true that f(0) = 0 .

(Dang it! [STRIKE]Halls[/STRIKE] just beat me to the punch.) DUH! Mark44 did. --- One of those guys with a GREEN name --- LOL!

Any quadratic function passing through (0, 0) has the form: f(x) = ax2 + bx + 0 .

Last edited: Sep 7, 2012
4. Sep 7, 2012

### Staff: Mentor

No, he didn't.:tongue:

5. Sep 7, 2012

### bugatti79

I dont understand why we need extra information because I managed to find the expression for when $(h,k)=(\frac{h_e}{2},1)$

Ie, we have $(0,0)$ and $(\frac{h_e}{2},1)$

$y=a(x-\frac{h_e}{2})^2+1 \implies a=\frac{-4}{h_e^2}$ when we use $(0,0)$

thus $\displaystyle y=\frac{-4}{h_e^2}(x-\frac{h_e}{2} )^2+1$

checking $(0,0)$ we get $f(0)=0$

For my latest problem I have just simply replaced $\frac{h_e}{2}$ with $\alpha h_e$....?

6. Sep 7, 2012

### Staff: Mentor

I'm confused - is it (h, k) = $(\alpha h_e, 1)$
or $(h,k)=(\frac{h_e}{2},1)$

This is not a difficult problem (it's precalculus), but the notation being used here gets in the way of understanding, IMO. a and $\alpha$ look a lot alike, and $h_{\epsilon}$ could be simplified to simply h. Possibly there's a reason for the subscript, but that reason is not relevant to what you're asking in this thread.

7. Sep 8, 2012

### bugatti79

$(h,k)=(\frac{h_e}{2},1)$ was just a different/another example to which I could find a quadratic expression along with $(0,0)$. Yes, we could drop the e have have it as h.

8. Sep 8, 2012

### vela

Staff Emeritus
How come this answer has $a$ and $h$ in it? It should be in terms of $\alpha$ and $h_e$, right?

9. Sep 8, 2012

### bugatti79

Yes you are correct.
I have managed to get the book answer by using 3 points on the graph along with $y(x)=ax^2+bx+c$. The 3 points are $(0,0), (\alpha h_e, 1), (h_e,0)$ Ie, 3 equations and 3 unknowns etc.

What I dont understand is why using $y(x)=a(x-h)^2+k$ does not work with 2 points $(0,0), (\alpha h_e,1)$ when this method works for $(0,0), (0.5h_e,1)$ as shown in the 2nd multi quote...

10. Sep 8, 2012

### vela

Staff Emeritus
In other words, the expression in the book is
$$y=\frac{1}{\alpha h_e^2(\alpha-1)}x^2 - \frac{1}{\alpha h_e (\alpha-1)}x.$$ You do realize that $h$ and $h_e$ aren't the same and $\alpha$ and $a$ aren't the same either, right? Did you think that writing down something that's inaccurate wouldn't cause confusion?

In any case, the problem is that either the book is wrong or you misunderstood the problem. If the book's answer is correct, the point $(\alpha h_e, 1)$ isn't generally the vertex of the parabola as you have claimed.

11. Sep 8, 2012

### SammyS

Staff Emeritus
If a quadratic function passes through the three points, $(0,\,0), (\alpha h_e,\, 1),\text{ and } (h_e,\,0)$, then the x coordinate of the vertex is $\displaystyle \frac{h_e}{2}\ .$ If in addition you are told that the vertex is at point $(\alpha h_e,\, 1)\,,$ then you know that $\displaystyle \alpha=\frac{1}{2}\ .$ You can also determine information about a, b, and c depending upon which situation holds.

* * * * * * * * * * * *

For the quadratic function, $y(x)=ax^2+bx+c\,:$

If this function passes through the origin, then c=0 .

The vertex of $y(x)=ax^2+bx+c\,,$ has an x-coordinate of $\displaystyle -\,\frac{b}{2a}\,,$ which we already identified as being $\displaystyle \frac{h_e}{2}\ .$ Therefore, $\displaystyle b=-ah_e\ .$

This function also passes through point $(\alpha h_e,\, 1)\,,$ so plugging that in gives the answer given in your text:
$\displaystyle y(x)=\frac{1}{\alpha h_e^2(\alpha-1)}x^2 - \frac{1}{\alpha h_e (\alpha-1)}x\ .$
after doing a little more algebra.

So, it turns out that the vertex is not necessarily at point $(\alpha h_e,\, 1)\ .$ That explains why you couldn't get the correct answer when you assumed that vertex.

12. Sep 9, 2012

### bugatti79

Aha, I see now what is happening. Thanks to all who posted on this thread.
Regards

13. Sep 10, 2012

### Staff: Mentor

bugatti79, please stop copying the problem statement. It was already copied in posts 2 and 3. There was no reason for you to copy it again, as you did in posts 5, 7, and 9. As I have mentioned before in other threads, all that extra LaTeX makes the thread load very slowly in my browser.