Question

How do I write a linear growth function that passes through two points, and an exponential growth function that will pass through one of the same points. The example I am using goes through points (-5,6) and (-8,31).

Can anyone tell me how to set this up? This is algebra 1.
Thank you in advance!

Answer 1

To find a linear growth function using two points, you can use the two given points to find the slope and then use the slope and one of your points to create the equation of the line.

Answer 2

Slope (m) is given by "rise"/"run", that is the vertical change divided by the horizontal change. Or
(y2 - y1)/(x2 - x1)

Answer 3

So let one of your points be (x1,y1) and the other be (x2,y2). It doesn't matter which.

Answer 4

If we let (-5,6) be (x1,y1) and (-8,31) be (x2,y2), then (y2 - y1)/(x2 - x1) = (31 - 6)/(-8-(-5)) which is 25/-13 or - 25/13

Answer 5

I did that and got 31-6/-8-(-5) = 25/13 = 1.92, so I am track with that, but it all falls apart after that.

Answer 6

I think we can get it back on the rails :D

Answer 7

Now, you have a couple of choices. You can use either the point-slope formula or the slope intercept form to come up with the equation of your line.

Answer 8

Do you have a preference?

Answer 9

slope intercept I think
Whichever is easier!

Answer 10

I think it may be easier to start with point-slope, then rearrange the equation to put it in slope intercept form.

Answer 11

I think the slope is 25/-3? -8-(-5)=-8+5=-3?

Answer 12

Yes :) You are correct.

Answer 13

I am addition challenged this morning!

Answer 14

I am so glad! This is part of a big project and I can't figure out this portion. It is so frustrating!

Answer 15

So we could use point slope: y - y1 = m(x - x1), since we have m and can use either point.

Answer 16

ok

Answer 17

Just plug them into the formula. Let's use the first point (-5,6)

Answer 18

So we have y - 6 = (-25/3)(x - (-5))

Answer 19

I am with you

Answer 20

And then we distribute the -25/3 on the right side: y - 6 = (-25/3)x -125/3

Answer 21

Making sense so far? They gave you ugly numbers to work with!

Answer 22

I know, my other set of numbers is even worse! I have to do this twice, so I put the easier #'s here, and then I can figure out the other set once I have an example.

Answer 23

Makes sense to me!

Answer 24

I had to make up a parabola and certain points, so that is where the #s came from.

Answer 25

Aaah

Answer 26

So now you do have the equation of a line in point slope form. If you want it in slope intercept (y = mx + b) form, you just need to isolate y. So then just add 6 to both sides.

Answer 27

Ok, any thoughts on an exponential. I watched a few videos, but can't figure it out. The lesson gives equations, but doesn't explain how to write them.

Answer 28

y = (-25/3)x - 125/3 + 6 and then combine the -125/3 + 6

Answer 29

Yes, for exponential you also use both points.

Answer 30

But here you start with the general equation: y = ab^x

Answer 31

And the trick is to figure out what a and b are.

Answer 32

So, you start by creating two equations using the points you were given. So, the first one would look like: 6 = ab^(-5)

Answer 33

And so the second one would be ...?

Answer 34

The second one is 31=ab^-8

Answer 35

Of course. :) And then the easiest thing to do is usually to find b first by eliminating a.

Answer 36

That is were I was having trouble.

Answer 37

So, one way you could do that is to divide the second equation by the first. In that way the a's will cancel out.

Answer 38

So, then 31/6 = (ab^-8)/(ab^-5)

Answer 39

What would that simplify to then?

Answer 40

The a's cancel and you have b^-8 over b^-5 on the right side?
Left is 31/6.

Answer 41

Yes. And then b^-8 divided by b^-5 simplifies to ...

Answer 42

I forgot the rule, trying to look back, the two negatives are confusing me.

Answer 43

That's ok. I won't make you look it up! When you divide, you subtract the exponents, so -8-(-5)

Answer 44

-3 again!

Answer 45

Yep!

Answer 46

So we have 31/6 = b^-3

Answer 47

Or 31/6 = 1/b^3

Answer 48

Because of the reciprocal property of negative exponents

Answer 49

right, I forgot about that!

Answer 50

So then, b^3 = 6/31

Answer 51

And then b will be the lovely cube root of 6/31 !!!

Answer 52

Which is just heinous, if you ask me!

Answer 53

ugh, I couldn't have picked worse numbers

Answer 54

Oh, I bet you could have, actually :D

Answer 55

That is part of why it was so hard to figure out, the samples all worked out to squares. So how do I work that into an exponential equation.

Answer 56

You just leave it as it is written. Put it in parentheses, then raise it to x.

Answer 57

One way you could write it is b = (6/31)^(1/3)

Answer 58

It has to make an exponential line with geogebra, so I am not sure if that format works.

Answer 59

So then when you raise it the the x power, you would have (6/31)^(1/3)*x

Answer 60

You know, my son used geogebra before

Answer 61

So I think we can get it formatted properly at the end, after we figure out what a is.

Answer 62

I think it needs to be a function, like f(x)=

Answer 63

Yes, I believe you are correct. So we just need to find our a. Then our function will be
f(x) = a*(6/31)^(x/3)

Answer 64

Ugh, it says illegal multiplication. Let me try it a different way.

Answer 65

What did you get for a?

Answer 66

I put it in with a? I don't know?

Answer 67

We need to get a first. I think that is why it said "illegal multiplication", since we need to put a number in for a.

Answer 68

So, now that we have b, we can use y = ab^x and one of the points to go back and find a.

Answer 69

It won't be a pretty number. ;)

Answer 70

Oh, this is supposed to be a fairly simple project, I have spent days on it!

Answer 71

We can use (-5,6) again, if you want?

Answer 72

Sure

Answer 73

I am sorry to hear that! I think you are right, the choice of points makes a big difference to how easy the problem is.

Answer 74

Generally speaking, if you can choose points where x = 0 and where x = 1

Answer 75

that will simplify your calculations

Answer 76

Do you have an image of the original question?

Answer 77

What you are doing should "work", it's just not the easiest way on account of the numbers.

Answer 78

I should all work, it just isn't easy, lol.

Answer 79

It not I

Answer 80

yes, hold on

Answer 81

LOL

Answer 82

I should work too, but this is me procrastinating! ;)

Answer 83

West Mathington’s most urgent need is a parabolic freeway. Create your own upward opening quadratic function, f(x), which has two real zeros. Prove that f(x) has two real zeros.

If you are doing this project collaboratively, then your partner will need to create a unique second freeway function, g(x), and prove that it also has two real zeros.
Two on-ramps are needed to be placed on the parabolic freeway. Decide where on the parabola of f(x) you are placing the on-ramp locations. Write those ordered pairs down.

If you are working collaboratively, identify four on-ramps, two on each function.
West Mathington wants to connect these on-ramps with some surface roads.
Create a linear growth function, h(x), that passes through both on-ramp points.
Create an exponential growth function, j(x), that passes through at least one of the on-ramp points.

Show all of the work you did to create both functions.

Collaborative pairs need to connect linear function h(x) through the on ramp points on f(x), and connect exponential function j(x) through any on ramp point on g(x). At least three of the four on ramp points will be used.
What important relationship does the x-coordinates of the on-ramp location points have with the system of equations formed by the two roads’ functions that are being connected? Provide justification and support for your explanation.

The city planner needs to identify the most northern road. Prove which road will eventually go the furthest to the north (positive y-direction). Create tables for your functions using an appropriate domain of five integers. Using the tables and graph, explain to the city planner which road will be the furthest north as the x values continue to get larger (the road continues to go east). Provide reasoning why.

Collaborative groups should prove whether f(x) or h(x) is the most northern between the two, and whether g(x) or j(x) is the most northern between those two.
Include your graph that shows the functions that model each of the roads and the on-ramps.

Answer 84

If you could upload an image of the original problem, we could probably find two different points and come up with linear and exponential equations quickly. Or, we can finish finding a for the points we were using and then go ahead and submit the resulting function.

Answer 85

I have to do it as a collaboration, so I did 2 parabolas, then picked 4 points, now on the part with linear growth for both on ramps and exponential growth.

Answer 86

The posts are out of order on my screen, do you see it?

Answer 87

I didn't want to redo the whole thing, but I am open to suggestions at this point. I don't know why I am having such a hard time with this.

Answer 88

Yes, it is very confusing!

Answer 89

Oh, it is way up at the top! Just a moment, please, and I'll read it. :)

Answer 90

Wow.

Answer 91

Can I ask you a question?

Answer 92

What class are you taking and in pursuit of what major? What book is this from?

Answer 93

I can tell you why you're having a hard time with it. It's a pretty wild problem!

Answer 94

Are we allowed to communicate outside of OpenStudy? I don't know what the "rules" are regarding privacy of users, but I think we could do this more easily by email!

Answer 95

Sorry my computer froze. IDK the rules, but If you want my email I am not sure how to do that without giving it to everyone

Answer 96

This is an FLVS honors algebra class. So yeah, 8th-9th grade!

Answer 97

Sorry, FLVS is Florida Virtual School.

Answer 98

I am trying to help my daughter with this. . .