Question

(a) Determine the polynomial function of least degree whose graph passes through the given points.

Answer 1

as we know given points are not in straight line so we discard 1 degree polynomial also with 4 points and quadratic might work

lets suppose it is of the form \(y=a x^{2}+b x+c\)

0,0

\(y=a x^{2}+b x+c\)

\(0=a 0+b 0+c\)

\(0=c\)

\(y=a x^{2}+b x+0\)

\(y=a x^{2}+b x\)

1,1

\(1=a+b\)

-1,7

\(7=a-b\)

solving we get \(a=4 b=-3\)

\(y=4 x^{2}-3 x\)

now it passes through 4,58

\(y=4 * 4^{2}-3 * 4=64-12=52\) which is not 58 hence it cant be quadratic

lets suppose it is of the form \(y=a x^{3}+b x^{2}+c x+d\)

0,0

\(0=a 0+b 0+c 0+d\)

\(\mathrm{d}=0\)

\(y=a x^{3}+b x^{2}+c x+0\)

\(y=a x^{3}+b x^{2}+c x\)

1,1

\(1=a+b+c\)

-1,7

\(7=-a+b-c\)

4,58

\(58=64 a+16 b+4 c\)

solve these

\(\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 7 \\ 64 & 16 & 4 & 58\end{array}\right]\)

\(R_{3}=R_{3}-64 R_{1}\)

\(\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 7 \\ 0 & -48 & -60 & -6\end{array}\right]\)

\(R_{2}=R_{2}+R_{1}\)

\(\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ 0 & 2 & 0 & 8 \\ 0 & -48 & -60 & -6\end{array}\right]\)

\(R_{3}=R_{3}+24 R_{2}\)

\(\left[\begin{array}{cccc}1 & 1 & 1 & 1 \\ 0 & 2 & 0 & 8 \\ 0 & 0 & -60 & 186\end{array}\right]\)

\(-60 c=186666\)

\(c=-3.1\)

\(2 b=8\)

\(b=4\)

\(a+b+c=1\)

\(a+4-3.1=1\)

\(a+0.9=1\)

\(a=0.1\)

\(\langle a, b, c>-><0.1,4,-3.1>\)

\(y=0.1 x^{3}+4 x^{2}-3.1 x\)

we can see first graph is correct

2)

let \(y=a x^{2}+b x+c\)

0,1

\(1=a 0+b 0+c\)

\(c=1\)

\(y=a x^{2}+b x+1\)

\(1 / 4,3 / 4\)

\(3 / 4=a / 16+b / 4+1\)

\(12=a+4 b+1\)

\(11-4 b=a\)

\(y=(11-4 b) x^{2}+b x+1\)

horizontal tangent meansat \(\mathrm{x}=1 / 4\) there is maxima or minima depending on sign of \(11-4 \mathrm{~b}\)

which occurs at \(x=-b / 2 a=-b / 2(11-4 b)=-b / 22-8 b=b / 8 b-22=1 / 4\)

\(4 b=8 b-22\)

\(4 b=22\)

\(b=5.5\)

\(a=11-4 b=-11\)

\(y=-11 x^{2}+5.5 x+1\)

hence 2 nd answer correct \(a<0\)and maxima at \(x=1 / 4\)