Find b and c so that the parabola http://webwork.rutgers.edu/webwork2_...bd38306ee1.png has vertex http://webwork.rutgers.edu/webwork2_...cbc6230861.png.

What do I do here?

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- Sep 21st 2011, 06:31 AMhabibixoxFinding the vertex at a given point?
Find b and c so that the parabola http://webwork.rutgers.edu/webwork2_...bd38306ee1.png has vertex http://webwork.rutgers.edu/webwork2_...cbc6230861.png.

What do I do here? - Sep 21st 2011, 06:34 AMSironRe: 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 AMSudharakaRe: Finding the vertex at a given point?
- Sep 21st 2011, 03:05 PMhabibixoxRe: Finding the vertex at a given point?
wait so is it

x^2-18x+91? - Sep 21st 2011, 03:07 PMhabibixoxRe: 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)

so final answer is 0? - Sep 21st 2011, 03:22 PMhabibixoxRe: 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 PMSudharakaRe: Finding the vertex at a given point?
- Sep 21st 2011, 10:29 PMSironRe: 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.