are my values off because my leading coefficent was not 1 in the second equation?
Your first equation you have a "b" which shouldn't be present and the 2nd equation you have no variables.
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...
c is a free variable.
The solution that takes into account ever possible solution written in terms of c.
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.
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....
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.
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