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Help me with this please please

Help me with this please please-example-1
User TMNuclear
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1 Answer

3 votes

Answer:

  1. f(x) = 2·5^x . . . b = 5; a = 2
  2. f(x) = (9/2)(2/3)^x . . . b = 2/3; a = 9/2
  3. f(x) = (1/2)4^x . . . b = 4; a = 1/2

Explanation:

Actually, the lesson is giving you wrong information. The exponential function written in the example does not work for the point (4, 24).

Here's another way to look at it.

Given

two points: (x1, y1) and (x2, y2)

Solution

The exponential function can be written as ...

y = (y1)(y2/y1)^((x -x1)/(x2 -x1))

Example

For the example points, the function would be ...

y = (6)(24/6)^((x-2)/(4-2)) = 6·4^(x/2 -1) = (6/4)(4^(1/2))^x = (3/2)2^x

Check

For x=2, y=(3/2)2^2 = 6. For x=4, y = (3/2)2^4 = 24.

You will notice that we figure out "b" and "a" as a part of the process of simplifying the equation. The equation would work perfectly well without simplification.

_____

1. Given points (2, 50) and (3, 250), the equation is ...

f(x) = 50(250/50)^((x -2)/(3 -2)) = 50·5^(x -2) = (50/25)5^x

f(x) = 2·5^x . . . b = 5; a = 2

__

2. Given points (1, 3) and (2, 9/2), the equation is ...

f(x) = (3)(3/(9/2))^((x -1)/(2 -1)) = (3)(2/3)^(x -1) = (3/(2/3))(2/3)^x

f(x) = (9/2)(2/3)^x . . . b = 2/3; a = 9/2

__

3. Given points (3, 32) and (4, 128), the equation is ...

f(x) = (32)(128/32)^((x -3)/(4 -3)) = (32)4^(x -3) = (32/64)4^x

f(x) = (1/2)4^x . . . b = 4; a = 1/2

_____

Additional comments

The usual rules of exponents are useful in this:

a^(b+c) = (a^b)(a^c)

a^(bc) = (a^b)^c

a^-b = 1/a^b

User Cowborg
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