Answer :
The given parabola passes through the points (1,4), (2, 8), and (4, 34)
The equation of the parabola is
[tex]y=ax^2+bx+c[/tex]Substitute x=1 and y=4, we get
[tex]4=a(1)^2+b(1)+c[/tex][tex]4=a+b+c[/tex][tex]a+b+c=4\text{ take it as equation (1).}[/tex]Substitute x=2 and y=8 in the parabola equation, we get
[tex]8=a(2)^2+b(2)+c[/tex][tex]8=4a+2b+c[/tex][tex]4a+2b+c=8\text{ take it as equation (2).}[/tex]Substitute x=4 and y=34 in the parabola equation, we get
[tex]34=a(4)^2+b(4)+c[/tex][tex]34=16a+4b+c[/tex][tex]16a+4b+c=34\text{ take it as equation (3)}[/tex]Hence we get the system of equations
[tex]a+b+c=4\text{ }[/tex][tex]4a+2b+c=8\text{ }[/tex][tex]16a+4b+c=34\text{ }[/tex]This can be written in the form AX=B, where
[tex]A=\begin{bmatrix}{1} & {1} & {1} \\ {4} & {2} & {1} \\ {16} & {4} & {1}\end{bmatrix}\text{ , X=}\begin{bmatrix}{} & {a} & {} \\ {} & {b} & {} \\ {} & {c} & {}\end{bmatrix}\text{ and }B=\begin{bmatrix}{} & {4} & {} \\ {} & {8} & {} \\ {} & {34} & {}\end{bmatrix}[/tex]The solution is
[tex]X=BA^{-1}[/tex]We need to find the inverse of the matrix A by using a calculator.
The inverse of the matrix A is
[tex]A^{-1}=\begin{bmatrix}{\frac{1}{3}} & {-\frac{1}{2}} & {\frac{1}{6}} \\ {-2} & {\frac{5}{2}} & {-\frac{1}{2}} \\ {\frac{8}{3}} & {-2} & {\frac{1}{3}}\end{bmatrix}[/tex]Multiplying inverse of A and B to find X.
[tex]\begin{bmatrix}{} & {a} & {} \\ {} & {b} & {} \\ {} & {c} & {}\end{bmatrix}=\begin{bmatrix}{} & {4} & {} \\ {} & {8} & {} \\ {} & {34} & {}\end{bmatrix}\begin{bmatrix}{\frac{1}{3}} & {-\frac{1}{2}} & {\frac{1}{6}} \\ {-2} & {\frac{5}{2}} & {-\frac{1}{2}} \\ {\frac{8}{3}} & {-2} & {\frac{1}{3}}\end{bmatrix}[/tex][tex]=\begin{bmatrix}{} & {4(\frac{1}{3})+8(-2)+34(\frac{8}{3})} & {} \\ {} & {4(-\frac{1}{2})+8(\frac{5}{2})+34(-2)} & {} \\ {} & {4(\frac{1}{6})+8(-\frac{1}{2})+34(\frac{1}{3})} & {}\end{bmatrix}[/tex][tex]=\begin{bmatrix}{} & {\frac{4}{3}-16+\frac{272}{3}} & {} \\ {} & {-2+20-68} & {} \\ {} & {\frac{2}{3}-4+\frac{34}{3}} & {}\end{bmatrix}[/tex][tex]=\begin{bmatrix}{} & {\frac{4-48+272}{3}} & {} \\ {} & {-50} & {} \\ {} & {\frac{2-12+34}{3}} & {}\end{bmatrix}[/tex][tex]=\begin{bmatrix}{} & {76} & {} \\ {} & {-50} & {} \\ {} & {8} & {}\end{bmatrix}[/tex]Hence the solution is
[tex]a=76,\text{ b=-50 and c=8}[/tex]Substitute a=76, b=-46 and c=8 in the parabola equation, we get
[tex]y=76x^2-50x+8[/tex]Hence the required parabola equation is
[tex]y=76x^2-50x+8[/tex]