I know how to find the equation of a cubic polynomial using 4 points and then a system of equations and/or a matrix but is it possible with only 3 points ? I have tried a matrix system but because one of the points is (zero, something) my matrix has a row of zeros and therefore is non invertible. Here is what I have:
A cubic polynomial is a function of the form
$$p(x)=ax^3+bx^2+cx+d$$
Find two cubic polynomials $$p_1(x)\ and\ p_2(x)\ such\ that$$
i. $$p_1(x) \ passes \ through\ the\ points\ (-2,1), (-1,0), (0,2)$$
ii. $$p_2(x) \ passes \ through \ the\ points\ (0,2), (1,0), (2,-2)$$
iii. $$p_1'(0)=p_2'(0)\ and \ p_1''(0)=p_2''(0)$$
I think I have to use the third statement but I am not sure how.
Thank you for your help
Best Answer
Let's start with $p_1(x)=a_1x^3+b_1x^2+c_1x+d_1$, $p_2(x)=a_2x^3+b_2x^2+c_2x+d_2$. We have that $p_1(0)=p_2(0)=2 \to d_1=d_2=2$. Next, $$p_1'(0)=p_2'(0) \to c_1=c_2=c$$ $$p_1''(0)=p_2''(0) \to b_1=b_2=b$$ We also have that $$p_1(-1)=0 \to -a_1+b-c+2=0$$ $$p_2(1)=0 \to a_2+b+c+2=0$$ $$p_1(-2)=1 \to -8a_1+4b-2c+2=1$$ $$p_2(2)=-2 \to 8a_2+4b+2c+2=-2$$ All that's left is to solve this linear system of 4 equations with 4 variables.