Chapter 10.3, Problem 61E

To determine

To find: The line, quadratic, cubic and fourth degree polynomial.

Answer to Problem 61E

  $\begin{array}{c}\text{Line: }y=12x\\ \text{Quadratic: }y=8x^2+4x\\ \text{Cubic: }y=-13x^3+47x^2-22x\\ \text{Fourth: }y=-4x^4+11x^3+3x^2+2x\end{array}$

  

Explanation of Solution

Given: The given points are $\left(0,0\right),\left(1,12\right),\left(2,40\right),\left(3,6\right),\left(-1,-14\right)$

The equation of line using $\left(0,0\right),\left(1,12\right)$

  $\begin{array}{c}\frac{y-12}{x-1}=\frac{12-0}{1-0}\\ y-12=12x-12\\ y=12x\left[\text{Equation of line}\right]\end{array}$

The equation of quadratic using first three points $\left(0,0\right),\left(1,12\right),\left(2,40\right)$

  $\begin{array}{l}y=a{x}^2+bx+c\\ \text{For point }\left(0,0\right)\\ c=0\mathrm{...}\left(i\right)\\ \text{For point }\left(1,12\right)\\ 12=a+b+c\mathrm{...}\left(ii\right)\\ \text{For point }\left(2,40\right)\\ 40=4a+2b+c\mathrm{...}\left(iii\right)\end{array}$

Using $\left(i\right),\left(ii\right)\text{ and }\left(iii\right)$

  $\begin{array}{l}a=8,b=4,c=0\\ y=8x^2+4x\left[\text{Equation of quadratic}\right]\end{array}$

The cubic equation using first four points $\left(0,0\right),\left(1,12\right),\left(2,40\right),\left(3,6\right)$

Let cubic equation be $y=a{x}^3+b{x}^2+cx+d$

  $\begin{array}{l}y=a{x}^3+b{x}^2+cx+d\\ \text{For point }\left(0,0\right)\\ d=0\mathrm{...}\left(i\right)\\ \text{For point }\left(1,12\right)\\ 12=a+b+c+d\mathrm{...}\left(ii\right)\\ \text{For point }\left(2,40\right)\\ 40=8a+4b+2c+d\mathrm{...}\left(iii\right)\\ \text{For point }\left(3,6\right)\\ 6=27a+9b+3c+d\mathrm{...}\left(iv\right)\end{array}$

Using $\left(i\right),\left(ii\right)\text{,}\left(iii\right)\text{ and }\left(iv\right)$

  $\begin{array}{l}a=-13,b=47,c=-22,d=0\\ y=-13x^3+47x^2-22x\left[\text{Equation of cubic}\right]\end{array}$

The fourth degree polynomial using first five points $\left(0,0\right),\left(1,12\right),\left(2,40\right),\left(3,6\right),\left(-1,-14\right)$

Let cubic equation be $y=a{x}^4+b{x}^3+c{x}^2+dx+e$

  $\begin{array}{l}y=a{x}^4+b{x}^3+c{x}^2+dx+e\\ \text{For point }\left(0,0\right)\\ e=0\mathrm{...}\left(i\right)\\ \text{For point }\left(1,12\right)\\ 12=a+b+c+d+e\mathrm{...}\left(ii\right)\\ \text{For point }\left(2,40\right)\\ 40=16a+8b+4c+2d+e\mathrm{...}\left(iii\right)\\ \text{For point }\left(3,6\right)\\ 6=81a+27b+9c+3d+e\mathrm{...}\left(iv\right)\\ \text{For point }\left(-1,-14\right)\\ -14=a-b+c-d+e\left(v\right)\end{array}$

Using $\left(i\right),\left(ii\right)\text{,}\left(iii\right)\text{,}\left(iv\right)\text{ and }\left(v\right)$

  $\begin{array}{l}a=-4,b=11,c=3,d=2\\ y=-4x^4+11x^3+3x^2+2x\left[\text{Equation of fourth degree}\right]\end{array}$

Using graphing utility graph all polynomial on same window.