# Construct quadratic equation using the vertex method

## Homework Statement

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##

## Homework Equations

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

## 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

Mark44
Mentor

## Homework Statement

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##

## Homework Equations

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

## 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

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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##

## Homework Equations

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

## 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 doesn't 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
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:
Mark44
Mentor
(Dang it! Halls just beat me to the punch.)
No, he didn't.:tongue:

## Homework Statement

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##

## Homework Equations

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

## 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

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.

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 .

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##....?

Mark44
Mentor
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##

## Homework Equations

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

## The Attempt at a Solution

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

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)##

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.

## Homework Statement

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##

## Homework Equations

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

## 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

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##....?

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.

##(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.

vela
Staff Emeritus
Homework Helper
##\displaystyle \frac{x^2}{((-1 + a) a h^2)}-\frac{x}{((-1 + a) a h)} ##
How come this answer has ##a## and ##h## in it? It should be in terms of ##\alpha## and ##h_e##, right?

## Homework Statement

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##

## Homework Equations

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

## 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

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##....?

##(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.

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

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...

vela
Staff Emeritus
Homework Helper
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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
...

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 don't 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...

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.

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.

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.

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

Mark44
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.