Question

(a) Make a scatter plot of the data in the table given to the right.(b) Find a power function that models the data.(c) Find a quadratic function that models the data.(d) Find a logarithmic function that models the data.

Answer 1

To find the power function or the exponential function that models the data, we will use the following standard formula:

`g(x)=a\cdot r^x`

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

`g(1)=a\cdot r^1=3.5`

And then, at x = 2, y = 5.7

`g(2)=ar^2=5.7`

From these 2 equations, let us try to solve for a.

From the first equation, we can rewrite it as

`r=(3.5)/(a)`

Substitute it to the second equation

`ar^2=5.7`
`a((3.5)/(a))^2=5.7`
`a((12.25)/(a^2)^{})=5.7`
`(12.25)/(a)^{}=5.7`
`12.25^{}=5.7a`
`a=2.149`

Now, to solve for r, we will just substitute a to the first equation

`r=(3.5)/(a)`
`r=(3.5)/(2.149)`
`r=1.629`

With these, we can now form a formula of:

`y=2.149(1.629^x)`

Now, to find the quadratic function the models the data, we will use the standard formula:

`y=Ax^2+Bx+C`

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

`y=Ax^2+Bx+C`
`3.5=A(1)^2+B(1)+C`
`A^{}+B+C=3.5`
`A^{}=3.5-B-C`

Next, substitute the second pair and the value of A and solve for B

`y=Ax^2+Bx+C`
`5.7=(3.5-B-C)(2)^2+B(2)+C`
`5.7=14-4B-4C+2B+C`
`5.7=14-2B-3C`
`2B=14-5.7-3C`
`2B=8.3-3C`
`B=4.15-1.5C`

Next, substitute the third pair and the values of A and B to the equation and the solve for C

`6.5=(3.5-(4.15-1.5C)-C)(3)^2+(4.15-1.5C)(3)+C`
`6.5=(3.5-4.15+1.5C-C)9+12.45-4.5C+C`
`6.5=31.5-37.35+13.5C-9C+12.45-4.5C+C`
`6.5=6.6+C`
`C=6.5-6.6`
`C=-0.1`

Since B = 4.15 - 1.5C

`B=4.15-1.5C`
`B=4.15-1.5(-0.1)_{}`
`B=4.15+0.15_{}`
`B=4.3`

Substitute B and C to solve for A

`A^{}=3.5-B-C`
`A^{}=3.5-4.3-(-0.1)`
`A^{}=3.5-4.3+0.1`
`A=-0.7`

Substitute the values of A, B, and C to the equation

`y=Ax^2+Bx+C`
`y=-.07x^2+4.3x-0.1`