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- May 11th 2010, 07:59 PMdwsmith
- May 11th 2010, 08:03 PMAri
are my values off because my leading coefficent was not 1 in the second equation?

- May 11th 2010, 08:04 PMdwsmith
Your first equation you have a "b" which shouldn't be present and the 2nd equation you have no variables.

- May 11th 2010, 08:06 PMAri
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... - May 11th 2010, 08:07 PMdwsmith
c is a free variable.

- May 11th 2010, 08:33 PMAri
- May 11th 2010, 08:34 PMdwsmith
The solution that takes into account ever possible solution written in terms of c.

- May 11th 2010, 08:35 PMAri
- May 11th 2010, 08:38 PMdwsmith
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. - May 11th 2010, 08:44 PMAri
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.... - May 11th 2010, 08:57 PMdwsmith
We have a minor mistake.

We lost a negative sign.

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.

Check. - May 11th 2010, 09:31 PMdwsmith
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 .

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

So our equation becomes:

Now we have

, and - May 12th 2010, 05:39 AMAri
- May 12th 2010, 05:55 AMdwsmith
- May 12th 2010, 06:08 AMAri