Question

How do you find the equation of each function? Function A is f(x)=3x^2 and Function C is f(x)=18x-16. How do you get those? And then how do you get the equation for Function B? Thanks!

Answer 1

`f(x) = 3^x` --- Function A

`f(x) = x^2 - 5` --- Function B

`f(x) = 18x -6` --- Function C

Explanation:

Solving (a): Equation of Function A

An exponential equation is represented as:

`y = ab^x`

From the graph of function A,

`x = 0\ when\ y = 1`

`x = 1\ when\ y = 3`

For:
`x = 0\ when\ y = 1`

`y = ab^x` becomes

`1 = ab^0`

`1 = a`

`a =1`

For:
`x = 1\ when\ y = 3`

`y = ab^x` becomes

`3 = a*b^1`

`3 = a*b`

Substitute 1 for a

`3 = 1*b`

`3 = b`

`b = 3`

`y = ab^x` becomes

`y = 1*3^x`

`y = 3^x`

Replace y with f(x)

`f(x) = 3^x`

Solving (b): Equation of Function B

A quadratic equation is represented as:

`y = ax^2 + bx + c`

From the table of function B,

`x = 3\ when\ y =4`

`x = 5\ when\ y =20`

`x = 6\ when\ y =31`

For:
`x = 3\ when\ y =4`

`y = ax^2 + bx + c` becomes

`4 = a*3^2 + b*3 + c`

`4 = a*9 + 3b + c`

`4 = 9a + 3b + c`

For
`x = 5\ when\ y =20`

`y = ax^2 + bx + c` becomes

`20 = a*5^2 + b*5 + c`

`20 = a*25 + b*5 + c`

`20 = 25a + 5b + c`

For
`x = 6\ when\ y =31`

`y = ax^2 + bx + c` becomes

`31 = a*6^2 + b*6 + c`

`31 = a*36 + b*6 + c`

`31 = 36a + 6b + c`

So, we solve for a, b and c in:

`4 = 9a + 3b + c`

`20 = 25a + 5b + c`

`31 = 36a + 6b + c`

Make c the subject in
`4 = 9a + 3b + c`

`c = 4 - 9a - 3b`

Substitute
`c = 4 - 9a - 3b` in
`20 = 25a + 5b + c` and
`31 = 36a + 6b + c`

`20 = 25a + 5b + c` becomes

`20 = 25a + 5b + 4-9a-3b`

Collect Like Terms

`-4+20 = 25a -9a+ 5b -3b`

`16 = 16a+ 2b`

Divide through by 2

`8 = 8a + b`

`c = 4 - 9a - 3b`

`31 = 36a + 6b + c` becomes

`31 = 36a + 6b + 4 - 9a - 3b`

Collect Like Terms

`-4+31 = 36a - 9a+ 6b - 3b`

`27 = 27a+ 3b`

Divide through by 3

`9 = 9a + b`

Solve for a and b in:
`8 = 8a + b` and
`9 = 9a + b`

Subtract both equations:

`8 - 9 = 8a - 9a + b - b`

`8 - 9 = 8a - 9a`

`-1 = -a`

Divide both sides by -1

`1 = a`

`a = 1`

Substitute 1 for a in
`9 = 9a + b`

`9 = 9*1 + b`

`9 = 9 +b`

Subtract 9 from both sides

`9-9=9-9+b`

`0=b`

`b = 0`

Substitute
`b = 0` and
`a = 1` in
`c = 4 - 9a - 3b`

`c = 4 - 9*1-3*0`

`c = 4 - 9-0`

`c = -5`

So, the equation is:

`y = ax^2 + bx + c`

`y=1*x^2 +0*x-5`

`y=x^2 -5`

Replace y with f(x)

`f(x) = x^2 - 5`

Solving (c): Equation of Function C

This is calculated using:

`f(x) =a + (x -1)d`

Where

`a = 12` -- the first term

d = the difference between successive terms

`d = 30 -12 = 48-30 = 66 -48`

`d =18`

So, we have:

`f(x) =a + (x -1)d`

`f(x) = 12 + (x - 1)*18`

Open bracket

`f(x) = 12 + 18x - 18`

Collect Like Terms

`f(x) = 18x - 18+12`

`f(x) = 18x -6`