Construct quadratic equation using the vertex method

In summary: In my opinion. I think you can ignore the subscript and just use h.The way I read the problem is that you're given two points on a parabola, the vertex (h, k) = ##(\alpha h_e, 1)## and the point (0, 0). You're to find the equation of this parabola.First, I would write the equation of the parabola like this:\displaystyle y = a(x - h)^2 + kThe reason for the change is that I want to emphasize that h and k are constants - they are the coordinates of the vertex - and that x and y are variables - they are the coordinates of any point on the par
  • #1
bugatti79
794
1

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
 
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  • #2
bugatti79 said:

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

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
bugatti79 said:

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 [itex]\alpha h_e[/itex] and the k is 1 .

So you have [itex]f(x)=a(x-\alpha h_e)^2 + 1\ .[/itex]

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

See where that leads you.

What else do you know about quadratic functions? Specifically, what do you know about the symmetry of quadratic functions?

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

Added in Edit:

Any quadratic function passing through (0, 0) has the form: f(x) = ax2 + bx + 0 .
 
Last edited:
  • #4
SammyS said:
(Dang it! Halls just beat me to the punch.)
No, he didn't.:tongue:
 
  • #5
bugatti79 said:

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

Mark44 said:
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 said:
The general expression for a quadratic function with vertex at (h, k) is
a(x - h)2 + k .​

In your case the h is [itex]\alpha h_e[/itex] and the k is 1 .

So you have [itex]f(x)=a(x-\alpha h_e)^2 + 1\ .[/itex]

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

See where that leads you.

What else do you know about quadratic functions? Specifically, what do you know about the symmetry of quadratic functions?

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

Added in Edit:

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


I don't 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
bugatti79 said:
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)##

bugatti79 said:
I don't 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.
 
  • #7
bugatti79 said:

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

bugatti79 said:
I don't 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 said:
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.
 
  • #8
bugatti79 said:
##\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?
 
  • #9
bugatti79 said:

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

bugatti79 said:
I don't 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##...?

bugatti79 said:
##(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 said:
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 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...
 
  • #10
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
bugatti79 said:
...

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, [itex](0,\,0), (\alpha h_e,\, 1),\text{ and } (h_e,\,0)[/itex], then the x coordinate of the vertex is [itex]\displaystyle \frac{h_e}{2}\ .[/itex] If in addition you are told that the vertex is at point [itex](\alpha h_e,\, 1)\,,[/itex] then you know that [itex]\displaystyle \alpha=\frac{1}{2}\ .[/itex] You can also determine information about a, b, and c depending upon which situation holds.

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

For the quadratic function, [itex]y(x)=ax^2+bx+c\,:[/itex]

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

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

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

So, it turns out that the vertex is not necessarily at point [itex](\alpha h_e,\, 1)\ .[/itex] That explains why you couldn't get the correct answer when you assumed that vertex.
 
  • #12
vela said:
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 said:
If a quadratic function passes through the three points, [itex](0,\,0), (\alpha h_e,\, 1),\text{ and } (h_e,\,0)[/itex], then the x coordinate of the vertex is [itex]\displaystyle \frac{h_e}{2}\ .[/itex] If in addition you are told that the vertex is at point [itex](\alpha h_e,\, 1)\,,[/itex] then you know that [itex]\displaystyle \alpha=\frac{1}{2}\ .[/itex] You can also determine information about a, b, and c depending upon which situation holds.

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

For the quadratic function, [itex]y(x)=ax^2+bx+c\,:[/itex]

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

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

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

So, it turns out that the vertex is not necessarily at point [itex](\alpha h_e,\, 1)\ .[/itex] 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
 
  • #13
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.
 

1. What is the vertex form of a quadratic equation?

The vertex form of a quadratic equation is y = a(x - h)^2 + k, where a is the coefficient of the x^2 term, h is the x-coordinate of the vertex, and k is the y-coordinate of the vertex.

2. How do I find the vertex of a quadratic equation using the vertex method?

To find the vertex of a quadratic equation using the vertex method, follow these steps:

  1. Write the equation in vertex form (y = a(x - h)^2 + k).
  2. The x-coordinate of the vertex is h, and the y-coordinate is k.

3. Can I use the vertex method to solve any quadratic equation?

Yes, the vertex method can be used to solve any quadratic equation. However, it may not always be the most efficient method, so it's important to be familiar with other methods such as factoring or the quadratic formula.

4. What is the benefit of using the vertex method to construct a quadratic equation?

The vertex method allows you to easily identify the vertex of a quadratic equation, which is important for understanding the shape and behavior of the graph. It also simplifies the process of finding the x-intercepts and y-intercept of the graph.

5. Can I use the vertex method to solve real-world problems?

Yes, the vertex method can be used to solve real-world problems that can be represented by a quadratic equation. For example, it can be used to find the optimal value in a business scenario or to determine the maximum height of a projectile.

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