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Question: How do you find the polynomial function whose graph passes through \(\left( 2,4 \right),\left( 3,6 \...

How do you find the polynomial function whose graph passes through (2,4),(3,6),(5,10)?\left( 2,4 \right),\left( 3,6 \right),\left( 5,10 \right)?

Explanation

Solution

We will use second polynomial fitting. Then we will find the corresponding coefficient matrix. Later, we will use the Cramer’s Rule to determine the unknowns.

Complete step by step solution:
We are given with three points, (2,4),(3,6)\left( 2,4 \right),\left( 3,6 \right) and (5,10).\left( 5,10 \right).
Therefore, we will use second polynomial fitting like y=ax2+bx+c.y=a{{x}^{2}}+bx+c.
We will get the following three equations:
When we consider the point (2,4),\left( 2,4 \right), we will get y1=ax12+bx1+c=a(2)2+b(2)+c.{{y}_{1}}=ax_{1}^{2}+b{{x}_{1}}+c=a{{\left( 2 \right)}^{2}}+b\left( 2 \right)+c.
When we consider the point (3,6),\left( 3,6 \right), we will get y2=ax22+bx2+c=a(3)2+b(3)+c.{{y}_{2}}=ax_{2}^{2}+b{{x}_{2}}+c=a{{\left( 3 \right)}^{2}}+b\left( 3 \right)+c.
When we consider the point (5,10),\left( 5,10 \right), we will get y3=ax32+bx3+c=a(5)2+b(5)+c.{{y}_{3}}=ax_{3}^{2}+b{{x}_{3}}+c=a{{\left( 5 \right)}^{2}}+b\left( 5 \right)+c.
These equations will give us the following system of equations:
4a+2b+c=4. 9a+3b+c=6. 25a+5b+c=10. \begin{aligned} & 4a+2b+c=4. \\\ & 9a+3b+c=6. \\\ & 25a+5b+c=10. \\\ \end{aligned}
Let us convert this system of equations into matrix form as AX=BAX=B where A=[421  931  2551  ],X=[a b c ]A=\left[ \begin{aligned} & \begin{matrix} 4 & 2 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 9 & 3 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 25 & 5 & 1 \\\ \end{matrix} \\\ \end{aligned} \right], X=\left[ \begin{matrix} a \\\ b \\\ c \\\ \end{matrix} \right] and B=[4 6 10 ].B=\left[ \begin{matrix} 4 \\\ 6 \\\ 10 \\\ \end{matrix} \right].
When we solve this using X=A−1B,X={{A}^{-1}}B, we will get the values the unknowns a,ba,b and c.c.
Here we are using Cremer’s rule to find the values of the unknowns a,ba,b and c.c.
Let us find the determinant of the matrix A.A.
⇒∣A∣=∣421  931  2551  ∣=4(3−5)−2(9−25)+1(45−75)=4×(−2)−2(−16)+(−30)\Rightarrow \left| A \right|=\left| \begin{aligned} & \begin{matrix} 4 & 2 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 9 & 3 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 25 & 5 & 1 \\\ \end{matrix} \\\ \end{aligned} \right|=4\left( 3-5 \right)-2\left( 9-25 \right)+1\left( 45-75 \right)=4\times \left( -2 \right)-2\left( -16 \right)+\left( -30 \right)
⇒∣A∣=−8+32−30=−8+2=−6.\Rightarrow \left| A \right|=-8+32-30=-8+2=-6.
Consider the matrix Ax1{{A}_{{{x}_{1}}}} which is constructed by replacing the first column of the matrix AA with the elements of the column matrix B.B.
So, the matrix Ax1{{A}_{{{x}_{1}}}} is given by Ax1=[421  631  1051  ].{{A}_{{{x}_{1}}}}=\left[ \begin{aligned} & \begin{matrix} 4 & 2 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 6 & 3 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 10 & 5 & 1 \\\ \end{matrix} \\\ \end{aligned} \right].
We are going to find the determinant of the matrix Ax1.{{A}_{{{x}_{1}}}}.
Therefore, ∣Ax1∣=4(3−5)−2(6−10)+1(30−30)=4×(−2)−2(−4)+0=−8+8=0.\left| {{A}_{{{x}_{1}}}} \right|=4\left( 3-5 \right)-2\left( 6-10 \right)+1\left( 30-30 \right)=4\times \left( -2 \right)-2\left( -4 \right)+0=-8+8=0.
Consider the matrix Ax2{{A}_{{{x}_{2}}}} which is constructed by replacing the second column of the matrix AA with the elements of the column matrix B.B.
So, the matrix Ax2=[441  961  25101  ].{{A}_{{{x}_{2}}}}=\left[ \begin{aligned} & \begin{matrix} 4 & 4 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 9 & 6 & 1 \\\ \end{matrix} \\\ & \begin{matrix} 25 & 10 & 1 \\\ \end{matrix} \\\ \end{aligned} \right].
As we have done earlier, we are going to find the determinant of the matrix Ay.{{A}_{y}}.
So, ∣Ax2∣=4(6−10)−4(9−25)+1(90−150)=4(−4)−4(−16)+(−60)=−16+64−60=−16+4=−12.\left| {{A}_{{{x}_{2}}}} \right|=4\left( 6-10 \right)-4\left( 9-25 \right)+1\left( 90-150 \right)=4\left( -4 \right)-4\left( -16 \right)+\left( -60 \right)=-16+64-60=-16+4=-12.
Consider the matrix Ax3{{A}_{{{x}_{3}}}} which is constructed by replacing the third column of the matrix AA with the elements of the column matrix B.B.
Thus, the matrix Ax3=[424  936  25510  ].{{A}_{{{x}_{3}}=}}\left[ \begin{aligned} & \begin{matrix} 4 & 2 & 4 \\\ \end{matrix} \\\ & \begin{matrix} 9 & 3 & 6 \\\ \end{matrix} \\\ & \begin{matrix} 25 & 5 & 10 \\\ \end{matrix} \\\ \end{aligned} \right].
The determinant of Ax3=4(30−30)−2(90−150)+4(45−75)=0−2(−60)+4(−30)=120−120=0.{{A}_{{{x}_{3}}}}=4\left( 30-30 \right)-2\left( 90-150 \right)+4\left( 45-75 \right)=0-2\left( -60 \right)+4\left( -30 \right)=120-120=0.
Now, the values of the unknowns can be obtained as follows:
a=∣Ax1∣∣A∣=0−6=0.a=\dfrac{\left| {{A}_{{{x}_{1}}}} \right|}{\left| A \right|}=\dfrac{0}{-6}=0.
b=∣Ax2∣∣A∣=−12−6=126=2.b=\dfrac{\left| {{A}_{{{x}_{2}}}} \right|}{\left| A \right|}=\dfrac{-12}{-6}=\dfrac{12}{6}=2.
c=∣Ax3∣∣A∣=0−6=0.c=\dfrac{\left| {{A}_{{{x}_{3}}}} \right|}{\left| A \right|}=\dfrac{0}{-6}=0.
Now we will substitute these values in any of the equations in the system of equations.
We will get,
4(0)+2(2)+0=4. 9(0)+3(2)+0=6. 25(0)+5(2)+0=10. \begin{aligned} & 4\left( 0 \right)+2\left( 2 \right)+0=4. \\\ & 9\left( 0 \right)+3\left( 2 \right)+0=6. \\\ & 25\left( 0 \right)+5\left( 2 \right)+0=10. \\\ \end{aligned}
We can see,
y1=0x12+2x1+0=2x1. y2=0x22+2x2+0=2x2. y3=0x32+2x3+0=2x3. \begin{aligned} & {{y}_{1}}=0x_{1}^{2}+2{{x}_{1}}+0=2{{x}_{1}}. \\\ & {{y}_{2}}=0x_{2}^{2}+2{{x}_{2}}+0=2{{x}_{2}}. \\\ & {{y}_{3}}=0x_{3}^{2}+2{{x}_{3}}+0=2{{x}_{3}}. \\\ \end{aligned}
Generally, y=2x.y=2x.

Hence the polynomial function whose graph passes through (2,4),(3,6)\left( 2,4 \right),\left( 3,6 \right) and (5,10)\left( 5,10 \right) y=2x.y=2x.

Note: In (2,4),x=2\left( 2,4 \right), x=2 and y=4.y=4. That is y=2⋅2=2x.y=2\cdot 2=2x. In (3,6),x=3\left( 3,6 \right), x=3 and y=6.y=6. That is, y=2⋅3=2x.y=2\cdot 3=2x. Similarly, in (5,10),x=5\left( 5,10 \right), x=5 and y=10.y=10. That is, y=2⋅5=2x.y=2\cdot 5=2x. Therefore the required function is y=2x.y=2x.
The determinant ∣a1a2a3  b1b2b3  c1c2c3  ∣=a1(b2c3−c2b3)−a2(b1c3−c1b3)+a3(b1c2−c1b2).\left| \begin{aligned} & \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\\ \end{matrix} \\\ & \begin{matrix} {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\\ \end{matrix} \\\ & \begin{matrix} {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\\ \end{matrix} \\\ \end{aligned} \right|={{a}_{1}}\left( {{b}_{2}}{{c}_{3}}-{{c}_{2}}{{b}_{3}} \right)-{{a}_{2}}\left( {{b}_{1}}{{c}_{3}}-{{c}_{1}}{{b}_{3}} \right)+{{a}_{3}}\left( {{b}_{1}}{{c}_{2}}-{{c}_{1}}{{b}_{2}} \right).