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]
