# Finding the vertex at a given point?

• Sep 21st 2011, 06:31 AM
habibixox
Finding the vertex at a given point?
• Sep 21st 2011, 06:34 AM
Siron
Re: Finding the vertex at a given point?
Use completing the square:
Convert $\displaystyle y$ to $\displaystyle y=(x-a)^2+q$
Where $\displaystyle (a,q)$ are the coordinates of the vertex.
• Sep 21st 2011, 06:39 AM
Sudharaka
Re: Finding the vertex at a given point?
Quote:

Originally Posted by habibixox

Hi habibixox,

At the vertex of the parabola the first derivative of y becomes zero. So calculate, $\displaystyle \frac{dy}{dx}$ and equate it to zero. Since the corresponding x value at the vertex is given you can find b. I hope you can do it now.
• Sep 21st 2011, 03:05 PM
habibixox
Re: Finding the vertex at a given point?
wait so is it
x^2-18x+91?
• Sep 21st 2011, 03:07 PM
habibixox
Re: Finding the vertex at a given point?
using the second method the derivative is
-8x + x right?
set that = 0

0= -8x + x
0= X(-8+1)
• Sep 21st 2011, 03:22 PM
habibixox
Re: Finding the vertex at a given point?
nvm b = 72 now im working on c ill get it :) i understand it...
• Sep 21st 2011, 06:18 PM
Sudharaka
Re: Finding the vertex at a given point?
Quote:

Originally Posted by habibixox
using the second method the derivative is
-8x + x right?
set that = 0

0= -8x + x
0= X(-8+1)

No.

$\displaystyle y=-4x^2+bx+c$

$\displaystyle \frac{dy}{dx}=-4(2x)+b(1)+0$

$\displaystyle \frac{dy}{dx}=-8x+b$

Hope you can continue.
• Sep 21st 2011, 10:29 PM
Siron
Re: Finding the vertex at a given point?
The method I was thinking about is the following.
Given is the parabola: $\displaystyle y=-4x^2+bx+c$ and the vertex point with coordinates $\displaystyle (9,10)$
In general we know we can directly read the coordinates of the vertex of a parabola if we convert $\displaystyle y$ to:
$\displaystyle y=a(x-h)^2+q$ where $\displaystyle (h,q)$ are the coordinates of the parabola.

First, divide every term by $\displaystyle -4$ so:
$\displaystyle \frac{y}{-4}=x^2-\frac{b}{4}x-\frac{c}{4}$
Use completing the square in the RHS:
$\displaystyle \frac{y}{-4}=\left(x-\frac{b}{8}\right)^2-\frac{b^2}{64}-\frac{c}{4}$
$\displaystyle \Leftrightarrow y=-4\left(x-\frac{b}{8}\right)^2+\frac{b^2}{16}+c$

That means:
(1)$\displaystyle \frac{b}{8}=9 \Leftrightarrow b=72$ (1)
(2)$\displaystyle \frac{b^2}{16}+c=10$ (2)
Substituting (1) in (2) gives:
$\displaystyle 324+c=10 \Leftrightarrow c=-314$

Therefore the parabola with vertex $\displaystyle (9,10)$ is:
$\displaystyle y=-4x^2+72x-314$

But as you see, the method sudharaka wants to use is easier.