Modelling a quadratic - Systems of linear equations

**dwsmith** · May 11th 2010, 06:59 PM

Originally Posted by Ari

okay so

a = 5 + b - c

b = -10 + 2

c = ?

$a=15-3c$
$b=10-2c$
$c$

$\begin{bmatrix} 15-3c\\ 10-2c\\ c \end{bmatrix}\rightarrow \begin{bmatrix} 15-3c\\ 10-2c\\ 3c \end{bmatrix}\rightarrow \begin{bmatrix} 15\\ 10\\ 0 \end{bmatrix}+c\begin{bmatrix} -3\\ -2\\ 1 \end{bmatrix}$

**Ari** · May 11th 2010, 07:03 PM

are my values off because my leading coefficent was not 1 in the second equation?

**dwsmith** · May 11th 2010, 07:04 PM

Your first equation you have a "b" which shouldn't be present and the 2nd equation you have no variables.

**Ari** · May 11th 2010, 07:06 PM

what is the coefficent for c then?

i am happy that i am not taking this course otherwise i would have for sure failed lol...

**dwsmith** · May 11th 2010, 07:07 PM

c is a free variable.

**Ari** · May 11th 2010, 07:33 PM

Originally Posted by dwsmith

$a=\frac{15}{3}-\frac{1}{3}c$
$b=\frac{-10}{3}+\frac{2}{3}c$
$c$

$\begin{bmatrix} \frac{15}{3}-\frac{1}{3}c\\ \frac{-10}{3}+\frac{2}{3}c\\ c \end{bmatrix}\rightarrow \begin{bmatrix} 15-c\\ -10+2c\\ 3c \end{bmatrix}\rightarrow \begin{bmatrix} 15\\ -10\\ 0 \end{bmatrix}+c\begin{bmatrix} -1\\ 2\\ 3 \end{bmatrix}$

then what is all this

**dwsmith** · May 11th 2010, 07:34 PM

The solution that takes into account ever possible solution written in terms of c.

**Ari** · May 11th 2010, 07:35 PM

Originally Posted by dwsmith

The solution that takes into account ever possible solution written in terms of c.

and so that quote is the final solutions... man next term is gonna suck.

**dwsmith** · May 11th 2010, 07:38 PM

Yes that is saying add the 1st column vector to c times the second column vector.
To avoid confusion of c, a=x, b=y, and c=z
If c=5, we have x=10, y=0, and z=15. This a solution. We can do this all day by choosing a c and then finding the values.

**Ari** · May 11th 2010, 07:44 PM

shouldnt there be definite values for a,b,c since P(x) = a^x2 + bx + c

and they pass through the point (-1,5) and have a horizontal tangent at (1,7)

im sure there must be definite values for a, b, and c. otherwise this function would not work....

**dwsmith** · May 11th 2010, 07:57 PM

We have a minor mistake.

We lost a negative sign.

$\begin{bmatrix} 1 & -1 & 1 & 5\\ -2 & 1 & 0 & 0 \end{bmatrix}$

I made all the corrections.

So to answer you question about the coefficients.

If c=0, we have (15,10,0) as coefficients.

We know (-1,5) exist.

$15x^2+10x+0c\rightarrow 15(-1)^2-10=5$

Check.

**dwsmith** · May 11th 2010, 08:31 PM

There is an error some where with how the derivative is affecting the linear system; however, I can't find it so I solved your equation a different way.

We know the slope of the tangent line is 0 at (1,7); therefore, the general equation is $\pm a(x-1)^2+7$ .

If the parabola opens up, (-1,5) wouldn't be a point on the parabola.

So our equation becomes: $-a(x-1)^2+7$

$-a(-1-1)^2+7=5\rightarrow -4a+7=5\rightarrow -4a=-2\rightarrow a=\frac{1}{2}$

Now we have $\frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$

$\frac{-1}{2}+x-4$

$a=\frac{-1}{2}$ , $b=1$ and $c=-4$

**Ari** · May 12th 2010, 04:39 AM

Originally Posted by dwsmith

There is an error some where with how the derivative is affecting the linear system; however, I can't find it so I solved your equation a different way.

We know the slope of the tangent line is 0 at (1,7); therefore, the general equation is $\pm a(x-1)^2+7$ .

If the parabola opens up, (-1,5) wouldn't be a point on the parabola.

So our equation becomes: $-a(x-1)^2+7$

$-a(-1-1)^2+7=5\rightarrow -4a+7=5\rightarrow -4a=-2\rightarrow a=\frac{1}{2}$

Now we have $\frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$

$\frac{-1}{2}+x-4$

$a=\frac{-1}{2}$ , $b=1$ and $c=-4$

i see, how did you get this part?
$\frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$

when you opened the brackets, you put the 7 in while leaving the negative half out?

would that make the answer
a=-1/2
b=1
c=6 and 1/2

**dwsmith** · May 12th 2010, 04:55 AM

Originally Posted by Ari

i see, how did you get this part?
$\frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$

when you opened the brackets, you put the 7 in while leaving the negative half out?

would that make the answer
a=-1/2
b=1
c=6 and 1/2

Being hasty and not wanting to deal with fractions.
Instead multiply through by 2
$-x^2+2x-1+14\rightarrow -x^2+2x+13$

**Ari** · May 12th 2010, 05:08 AM

Originally Posted by dwsmith

Being hasty and not wanting to deal with fractions.
Instead multiply through by 2
$-x^2+2x-1+14\rightarrow -x^2+2x+13$

wow thanks very much!

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