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:

[tex]f(x) = 3^x[/tex] --- Function A

[tex]f(x) = x^2 - 5[/tex] --- Function B

[tex]f(x) = 18x -6[/tex] --- Function C

Step-by-step explanation:

Solving (a): Equation of Function A

An exponential equation is represented as:

[tex]y = ab^x[/tex]

From the graph of function A,

[tex]x = 0\ when\ y = 1[/tex]

[tex]x = 1\ when\ y = 3[/tex]

For: [tex]x = 0\ when\ y = 1[/tex]

[tex]y = ab^x[/tex] becomes

[tex]1 = ab^0[/tex]

[tex]1 = a[/tex]

[tex]a =1[/tex]

For: [tex]x = 1\ when\ y = 3[/tex]

[tex]y = ab^x[/tex] becomes

[tex]3 = a*b^1[/tex]

[tex]3 = a*b[/tex]

Substitute 1 for a

[tex]3 = 1*b[/tex]

[tex]3 = b[/tex]

[tex]b = 3[/tex]

[tex]y = ab^x[/tex] becomes

[tex]y = 1*3^x[/tex]

[tex]y = 3^x[/tex]

Replace y with f(x)

[tex]f(x) = 3^x[/tex]

Solving (b): Equation of Function B

A quadratic equation is represented as:

[tex]y = ax^2 + bx + c[/tex]

From the table of function B,

[tex]x = 3\ when\ y =4[/tex]

[tex]x = 5\ when\ y =20[/tex]

[tex]x = 6\ when\ y =31[/tex]

For: [tex]x = 3\ when\ y =4[/tex]

[tex]y = ax^2 + bx + c[/tex] becomes

[tex]4 = a*3^2 + b*3 + c[/tex]

[tex]4 = a*9 + 3b + c[/tex]

[tex]4 = 9a + 3b + c[/tex]

For [tex]x = 5\ when\ y =20[/tex]

[tex]y = ax^2 + bx + c[/tex] becomes

[tex]20 = a*5^2 + b*5 + c[/tex]

[tex]20 = a*25 + b*5 + c[/tex]

[tex]20 = 25a + 5b + c[/tex]

For [tex]x = 6\ when\ y =31[/tex]

[tex]y = ax^2 + bx + c[/tex] becomes

[tex]31 = a*6^2 + b*6 + c[/tex]

[tex]31 = a*36 + b*6 + c[/tex]

[tex]31 = 36a + 6b + c[/tex]

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

[tex]4 = 9a + 3b + c[/tex]

[tex]20 = 25a + 5b + c[/tex]

[tex]31 = 36a + 6b + c[/tex]

Make c the subject in [tex]4 = 9a + 3b + c[/tex]

[tex]c = 4 - 9a - 3b[/tex]

Substitute [tex]c = 4 - 9a - 3b[/tex] in [tex]20 = 25a + 5b + c[/tex] and [tex]31 = 36a + 6b + c[/tex]

[tex]20 = 25a + 5b + c[/tex] becomes

[tex]20 = 25a + 5b + 4-9a-3b[/tex]

Collect Like Terms

[tex]-4+20 = 25a -9a+ 5b -3b[/tex]

[tex]16 = 16a+ 2b[/tex]

Divide through by 2

[tex]8 = 8a + b[/tex]

[tex]c = 4 - 9a - 3b[/tex]

[tex]31 = 36a + 6b + c[/tex] becomes

[tex]31 = 36a + 6b + 4 - 9a - 3b[/tex]

Collect Like Terms

[tex]-4+31 = 36a - 9a+ 6b - 3b[/tex]

[tex]27 = 27a+ 3b[/tex]

Divide through by 3

[tex]9 = 9a + b[/tex]

Solve for a and b in: [tex]8 = 8a + b[/tex] and [tex]9 = 9a + b[/tex]

Subtract both equations:

[tex]8 - 9 = 8a - 9a + b - b[/tex]

[tex]8 - 9 = 8a - 9a[/tex]

[tex]-1 = -a[/tex]

Divide both sides by -1

[tex]1 = a[/tex]

[tex]a = 1[/tex]

Substitute 1 for a in [tex]9 = 9a + b[/tex]

[tex]9 = 9*1 + b[/tex]

[tex]9 = 9 +b[/tex]

Subtract 9 from both sides

[tex]9-9=9-9+b[/tex]

[tex]0=b[/tex]

[tex]b = 0[/tex]

Substitute [tex]b = 0[/tex] and [tex]a = 1[/tex] in [tex]c = 4 - 9a - 3b[/tex]

[tex]c = 4 - 9*1-3*0[/tex]

[tex]c = 4 - 9-0[/tex]

[tex]c = -5[/tex]

So, the equation is:

[tex]y = ax^2 + bx + c[/tex]

[tex]y=1*x^2 +0*x-5[/tex]

[tex]y=x^2 -5[/tex]

Replace y with f(x)

[tex]f(x) = x^2 - 5[/tex]

Solving (c): Equation of Function C

This is calculated using:

[tex]f(x) =a + (x -1)d[/tex]

Where

[tex]a = 12[/tex] -- the first term

d = the difference between successive terms

[tex]d = 30 -12 = 48-30 = 66 -48[/tex]

[tex]d =18[/tex]

So, we have:

[tex]f(x) =a + (x -1)d[/tex]

[tex]f(x) = 12 + (x - 1)*18[/tex]

Open bracket

[tex]f(x) = 12 + 18x - 18[/tex]

Collect Like Terms

[tex]f(x) = 18x - 18+12[/tex]

[tex]f(x) = 18x -6[/tex]

-6x-7=4x-2
______________________________

PLEASE I BEG HELP PLEASE WILL MARK BRAINLIEST

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