We have:
[MATH]y(x)=Ax^2+Bx+C[/MATH]
Thus:
[MATH]y'(x)=2Ax+B[/MATH]
Now, from the two given points, we have:
[MATH]y_1=Ax_1^2+Bx_1+C[/MATH]
[MATH]y_2=Ax_2^2+Bx_2+C[/MATH]
And from the two given slopes, we have:
[MATH]m_1=2Ax_1+B[/MATH]
[MATH]m_2=2Ax_2+B[/MATH]
Using these last 2 equations, if we subtract the former from the latter, we will eliminate
B:
[MATH]m_2-m_1=2A\left(x_2-x_1\right)[/MATH]
Solve this for
A, and then substitute for
A into either of those equations whose difference we found to get
B. Then using either of the two equations that arose from the points on the quadratic, substitute for
A and
B to get
C.
