[MOVED] Find expo. fcn a b^(kx) + c passing through (2, 0),

[xomandi]

1) Find an exponential function of the form f(x) = a b^(kx) + c that goes through the points (2, 0), (5, 0), and (8,0) or within 0.1 of doing so. The value of b cannot be 1, a cannot be 0, and k cannot be 0.

I came up with f(x)= 4^-x. This graph has a horizontal asymptote at y = 0 so my graph goes within 0.1 of my designated points. But I have no + c, so I guess my solution must be wrong. Can I get some help with this?

2) Find a function that either goes through each of the points (-2,1), (1,3), and (4,0) exactly or within 0.1 of doing so.

It obviously can't be linear. I can see we have one zero given to us where x = 4. So (x - 4) is a factor of the function I am looking for. But other than that I am completely lost.

By the way: We are not supposed to use our regression applications. Ahhh

Thanks,
Mandi

 

[stapel]

1) You have f(x) = a b^kx + c, and the three points (2, 0), (5, 0), and (8, 0). Since b is not 1 and k is not 0, then this is a true exponential; that is, we cannot assign 1 = b or k = 0 and then draw a horizontal line along the x-axis.

Since the form of f(x) is exponential, it must then grow or decay; that's what exponentials do. The three points are in an exact straight line, which doesn't fit. Fortunately, we're given the 0.1 "wiggle room". So we can pick new points and then solve that system.

For instance, we could decide that f(x) is growing, and pick the points to be:

. . . . .(2, -0.97)
. . . . .(5, 0)
. . . . .(8, 0.03)

Why did I pick those? Well, 2 is three less than 5, and 5 is three less than 8, so counting by 3's, in some manner, seemed intelligent. Also, by decrementing and incrementing the endpoints by 0.03, we're within our margin of error of 0.1.

Start with the easy point, (5, 0):

. . . . .f(5) = a b^5k + c = 0

Then:

. . . . .a b^5k = -c

. . . . .b^5k = -c/a

If k = 1 (not required, but simple), then we get:

. . . . .b⁵ = -c/a

So, whatever b is, we need -c/a to be the fifth power of b. Exponentials usually (but not always) have a positive base, so b⁵ will be positive, and thus one of a and c needs to be negative, in order for -c/a to equal the positive b⁵.

Suppose b = 2. Then b⁵ = 32. Then -c/a will have to equal 32 as well.

Let's see what we can get from the other two points:

. . . . .f(2) = 4a + c = -0.97

. . . . .f(8) = 256a + c = 0.03

This is a system of two equations in two variables, which can be solved exactly.

Note: This is only one of many, many valid solutions.

2) Through any three non-linear points, you should be able to put a quadratic. So plug the three points into "y = ax² + bx + c", and solve the resulting system of equations.

There are other options, as well, but the above is probably the simplest, and will certainly be exact.

If you get stuck, please reply showing what you have tried and how far you have gotten. Thank you.

Eliz.

 

[xomandi]

gah I understood everything up to this part:

"Let's see what we can get from the other two points:

. . . . .f(2) = 4a + c = -0.97

. . . . .f(8) = 256a + c = 0.03

This is a system of two equations in two variables, which can be solved exactly. "

can you explain further??

 

[xomandi]

okay for the second one.

i plugged in my x and y values into the equation.

(4,0):

0 = a(4)^2 + b(4) + c

0 = 16a + 4b + c

c = -16a - 4b

(-2,1)

1 = a(-2)^2 + b(-2) + c

1 = 4a -2b + c

-4a = -2b + c - 1

a = 1/2b - 1/4c + 1/4

(1,3):

3 = a(1)^2 + b(1) + c

3 = a + b + c

b= -a - c +3


now i have to plug them into each other.. correct?

i havent done this in a while.

so i did alot of substituting and i now have a= 2/9b - 2/9
c = 5a + 3

b = -a - (5a + 3) + 3
b = -6a - 6

b = -6(2/9b - 2/9)
b = -4/3b + 4/3 - 6
7/3b = -14/3
b = -2

blahhhhh i got a = 2/9 and c = 37/9

when i graphed it, the parabola didnt go through my original points.

im so frustrated

 

[stapel]

can you explain further?

Which part? The plugging-in of the x-values? The evaluating of the function? Or the solving of a linear system?

(Since this sort of exercise generally pre-supposes knowledge of linear systems, I assumed you were familiar with this topic. If not, then we may have a problem.)

Thank you.

Eliz.

 

[xomandi]

i dont see where you got the 4 and the 256

edit// WAIT okayyy so you were using b = 2.. okay sorry i got confused

shouldnt it be -0.03 instead of -0.97??

 

[stapel]

shouldnt it be -0.03 instead of -0.97??

Yes; that would make more sense. (I'm afraid that post kind of got away from me....)

Eliz.

 

[xomandi]

okay so when i plugged everything in I got a = 0.06/252
and c = -0.031

when I plugged it into my y= for some reason it starts as an exponential decay and goes right into a straight line at -0.031.. should i be concerned about that?

also can you look at my post about problem number 2 and tell me if I was on the right track?? The solutions I got for that didnt match up when I graphed it.

 

[xomandi]

AHHH i just got an e-mail from my teacher saying that number 2 can be a piecewise function!

2) Find a function that either goes through each of the points (-2,1), (1,3), and (4,0) exactly or within 0.1 of doing so.