Sketch the spline of degree 2 with value 0.5 at x = 2.5 and the values 1, 1, 0, 0 at t0, . . . , t3, respectively. (t0=0, t1=2, t2=3, and t3=5) What is the value of the spline at x = 1 and 4? What I don't understand is how to find both the quadratics shown for 0
Suppose the first quadratic is $p(x)=ax^2+bx+c$. That implies that $p'(x)=2ax+b$. You require $p(0)=p(2)=1$ and $p'(2)=-1$, so $c=1,4a+2b+c=1$ and $4a+b=-1$. Hence $a=-\frac{1}{2},b=1,c=1$ and the quadratic is $1+\frac{1}{2}x(2-x)$. An exactly similar approach works for the second quadratic.