1. Originally Posted by Ari
okay so

a = 5 + b - c

b = -10 + 2

c = ?

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

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

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

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

4. 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...

5. c is a free variable.

6. 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

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

8. 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.

9. 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.

10. 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....

11. We have a minor mistake.

We lost a negative sign.

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

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.

12. 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 $c=-4$

13. 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?

a=-1/2
b=1
c=6 and 1/2

14. 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?

a=-1/2
b=1
c=6 and 1/2
Being hasty and not wanting to deal with fractions.
$-x^2+2x-1+14\rightarrow -x^2+2x+13$
$-x^2+2x-1+14\rightarrow -x^2+2x+13$