To find the power function or the exponential function that models the data, we will use the following standard formula:
[tex]g(x)=a\cdot r^x[/tex]From the table, assuming that g(x) = y, we can solve for the value of a and r.
As we can see from the table, when x = 1, y is 3.5
Substitute it from our equation
[tex]g(1)=a\cdot r^1=3.5[/tex]And then, at x = 2, y = 5.7
[tex]g(2)=ar^2=5.7[/tex]From these 2 equations, let us try to solve for a.
From the first equation, we can rewrite it as
[tex]r=\frac{3.5}{a}[/tex]Substitute it to the second equation
[tex]ar^2=5.7[/tex][tex]a(\frac{3.5}{a})^2=5.7[/tex][tex]a(\frac{12.25}{a^2}^{})=5.7[/tex][tex]\frac{12.25}{a}^{}=5.7[/tex][tex]12.25^{}=5.7a[/tex][tex]a=2.149[/tex]Now, to solve for r, we will just substitute a to the first equation
[tex]r=\frac{3.5}{a}[/tex][tex]r=\frac{3.5}{2.149}[/tex][tex]r=1.629[/tex]With these, we can now form a formula of:
[tex]y=2.149(1.629^x)[/tex]Now, to find the quadratic function the models the data, we will use the standard formula:
[tex]y=Ax^2+Bx+C[/tex]Here, we will use three ordered pairs from the given table
( 1 , 3.5 )
( 2 , 5.7 )
( 3 , 6.5 )
Substitute the first pair to the equation, and solve for A
[tex]y=Ax^2+Bx+C[/tex][tex]3.5=A(1)^2+B(1)+C[/tex][tex]A^{}+B+C=3.5[/tex][tex]A^{}=3.5-B-C[/tex]Next, substitute the second pair and the value of A and solve for B
[tex]y=Ax^2+Bx+C[/tex][tex]5.7=(3.5-B-C)(2)^2+B(2)+C[/tex][tex]5.7=14-4B-4C+2B+C[/tex][tex]5.7=14-2B-3C[/tex][tex]2B=14-5.7-3C[/tex][tex]2B=8.3-3C[/tex][tex]B=4.15-1.5C[/tex]Next, substitute the third pair and the values of A and B to the equation and the solve for C
[tex]6.5=(3.5-(4.15-1.5C)-C)(3)^2+(4.15-1.5C)(3)+C[/tex][tex]6.5=(3.5-4.15+1.5C-C)9+12.45-4.5C+C[/tex][tex]6.5=31.5-37.35+13.5C-9C+12.45-4.5C+C[/tex][tex]6.5=6.6+C[/tex][tex]C=6.5-6.6[/tex][tex]C=-0.1[/tex]Since B = 4.15 - 1.5C
[tex]B=4.15-1.5C[/tex][tex]B=4.15-1.5(-0.1)_{}[/tex][tex]B=4.15+0.15_{}[/tex][tex]B=4.3[/tex]Substitute B and C to solve for A
[tex]A^{}=3.5-B-C[/tex][tex]A^{}=3.5-4.3-(-0.1)[/tex][tex]A^{}=3.5-4.3+0.1[/tex][tex]A=-0.7[/tex]Substitute the values of A, B, and C to the equation
[tex]y=Ax^2+Bx+C[/tex][tex]y=-.07x^2+4.3x-0.1[/tex]