Curve Fitting Find the function $f\left(x\right)=a{x}^3+b{x}^2+cx+d$for which $f(-2)=-10,f(-1)=3,f\left(1\right)=5$, and $f\left(3\right)=15$.

We have been given:

$f(-2)=-10,f(-1)=3,f\left(1\right)=5$and$f\left(3\right)=15$

We have to find role="math" localid="1647455224381" $f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

$$\begin{gathered} a\cdot (-2{)}^3+b\cdot (-2{)}^2+c\cdot (-2)+d=-10 \\ -8a+4b-2c+d=-10 \end{gathered}$$

$$\begin{gathered} a\cdot (-1{)}^3+b\cdot (-1{)}^2+c\cdot (-1)+d=3 \\ -a+b-c+d=3 \end{gathered}$$

$$\begin{gathered} a\cdot (1{)}^3+b\cdot (1{)}^2+c\cdot \left(1\right)+d=5 \\ a+b+c+d=5 \end{gathered}$$

$$\begin{gathered} a\cdot (3{)}^3+b\cdot (3{)}^2+c\cdot \left(3\right)+d=15 \\ 27a+9b+3c+d=15 \end{gathered}$$

$$\begin{gathered} \left\{\begin{array}{l}-8a+4b-2c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-10\\ -a+b-c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=3\\ a+b+c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5\\ 27a+9b+3c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=15\end{array}\right. \\ \mathrm{The}\mathrm{corresponding}\mathrm{augmented}\mathrm{matrix}\mathrm{is}: \\ \left[\begin{array}{ccccc}-8& 4& -2& 1& -10\\ -1& 1& -1& 1& 3\\ 1& 1& 1& 1& 5\\ 27& 9& 3& 1& 15\end{array}\right] \end{gathered}$$

$$\begin{gathered} R_1\leftrightarrow R_3 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ -1& 1& -1& 1& 3\\ -8& 4& -2& 1& -10\\ 27& 9& 3& 1& 15\end{array}\right] \\ {R}_2\to R_2+R_1; \\ {R}_3\to R_3+8R_1; \\ {R}_4\to R_4-27R_1 \\ \left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 2& 0& 2& 8\\ 0& 12& 6& 9& 30\\ 0& -18& -24& -26& -120\end{array}\right] \\ {R}_3\to R_3-6R_2; \\ {R}_4\to R_4+9R_2 \\ \begin{array}{r}:\left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 2& 0& 2& 8\\ 0& 0& 6& -3& -18\\ 0& 0& -24& -8& -48\end{array}\right]\end{array} \end{gathered}$$

$$\begin{gathered} R_2\to \frac{1}{2}{R}_2; \\ {R}_4\to R_4+4R_3 \\ \begin{array}{r}:\left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 1& 0& 1& 4\\ 0& 0& 6& -3& -18\\ 0& 0& 0& -20& -120\end{array}\right]\end{array} \\ {R}_3\to \frac{1}{3}{R}_3; \\ {R}_4\to -\frac{1}{20}{R}_4 \\ :\left[\begin{array}{ccccc}1& 1& 1& 1& 5\\ 0& 1& 0& 1& 4\\ 0& 0& 2& -1& -6\\ 0& 0& 0& 1& 6\end{array}\right] \end{gathered}$$

$$\begin{gathered} \left\{\begin{array}{l}a+b+c+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=5\\ b+d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=4\\ 2c-d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-6\\ d\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=6\end{array}\right. \\ d=6 \\ \left(3\right):2c-6=-6 \\ c=0 \\ \left(2\right):b+6=4 \\ b=-2 \\ \left(1\right):a-2+6=5 \\ a=1 \end{gathered}$$

Therefore, the function will be $f\left(x\right)=x^3-2x^2+6$.