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Given the points (2,6) and (4,12) determine the (a) linear function, (b) exponential function and (c) power function they determine.

(a) $3 x$(b) $3(2)^{x / 2} \approx 3(1.414)^{x}$(c) $3 x$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 2

Exponential Functions

Harvey Mudd College

Baylor University

Idaho State University

Lectures

05:04

Given the points (1,5) and…

01:54

For the following exercise…

11:30

02:08

Write an exponential funct…

02:04

Finding an Exponential Fun…

01:39

03:14

02:40

03:56

Find a formula for an expo…

04:08

Find the indicated functio…

01:27

Write an exponential equat…

01:29

So if we want to determine what, um, Equation would best represent these points given a line exponential and a power function, we can first just go ahead and solve for these, like you normally would and then kind of go from there. So let's first go ahead and sulfur are slope of our line. So that's going to be y tu minus y one over x two minus x one. Um, so I'm just going to say these are the second points. This is the first. Should be 12/4 and then minus, uh 6/2. So that would be 6/2, which gives us three. And so now if we go ahead and use the point slope form would be why myself I one as referred to in times X mice X one. So it was 26 so it would be y minus six is equal to three x minus two. I'm going to add the six over, So B Y is in 23 x minus six plus six. But those council, so just get y is in 23 x. Okay, so this is our line. But you might also notice that This is in the exact same form as our power function over here. So we actually already have our power function as well. It's just going to be Why is ego to three X? So we actually don't even need to do anything for this because we already have it over there. Um, now we can go ahead and just plug those points into this. So let's see what we get, right? So if I first plug into six, that's going to give six is equal to a to the B squared, and then if I plug that next one in, it would be 12 is equal to a B to the fourth. Now, what I want to do is divi each of these equations. So over here we get one half is equal to, um So the A's cancel. And then we'd have B squared over B to the fourth, which is going to be be to the negative second, once we subtract the powers, um, and then we can raise each side to the negative one half power. So that gives us B is equal to one half raised to the negative one half. But the negative reciprocates it, and then the one half is just a square roots, this is going to be B is equal to route to. Okay, we have that. And now for us to solve for a we can just go ahead and plug this into one of those. So I'll plug it into this top equation. So we have six is equal to a route two squared, so root two squared is to divide over. That's a is even 23 So we end up with why is equal to three times Route two raised to the X. So that is going to be our exponential. Then, actually, let me go ahead and just write my power function down here as well. Yeah, so those will be our three equations that would best defy that data.

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