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

Answer:

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

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

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

Step-by-step 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

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

$f(x) = 18x -6$