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- #26

- Thread starter needalgebra
- Start date

- Admin
- #26

- Thread starter
- #27

- Aug 13, 2013

- 45

i have no diea how to do that, do you mind showing me how?

- Admin
- #28

I recommend you put the functions into your graphing calculator, and zoom in on the point of intersection until you have the desired accuracy.

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- #29

- Aug 13, 2013

- 45

I got this....

36.9999 = y

42 = x

36.9999 = y

42 = x

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- #30

seems pretty close to me, particularly if we can be satisfied with integral coordinates.

I "tricked" W|A into giving closer approximations:

\(\displaystyle (x,y)\approx(41.97171992699835,36.60518626378525)\)

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- #31

- Aug 13, 2013

- 45

I have one more graph to do.

I have to do it without the included gigabytes.

Here is what i got:

Quadratic model : f(x) 1/82 x2

domain: x>0

Logarithmic model : f(x) (30 - 1/2ln (26/25) Ln(x)

domain: no idea.

looking forward to your repsonse!

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- #32

For the quadratic model, you want:

\(\displaystyle f(x)=ax^2+b\)

where:

\(\displaystyle f(0)=a(0)^2+b=b=20\)

\(\displaystyle f(1)=a(1)^2+b=a+b=20.5\)

I would solve by substitution.

For the logarithmic model, we will have to pick a value with which to shift the graph to the left since $\ln(0)$ is undefined.

I would choose:

\(\displaystyle f(x)=a+b\ln(x+1)\)

where:

\(\displaystyle f(0)=a+b\ln(0+1)=a=30\)

\(\displaystyle f(1)=a+b\ln(1+1)=a+b\ln(2)=30.5\)

I would use substitution here as well.

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- #33

- Aug 13, 2013

- 45

1) f(x) = 0.5x^2 + 20

2) f(x)=30+0.72[ln(x+1)

1) f(x)=ax2+b

f(0)=a(0)2+b=b=20

f(1)=a(1)2+b=a+b=20.5 => a = 20.5 - 20 = 0.5 so,

f(x)=ax2+b f(x) = 0.5x^2 + 20

2) f(x)=a+bln(x+1)

f(0)=a+bln(0+1)=a=30

f(1)=a+bln(1+1)=a+bln(2)=30.5 =>b = (30.5-30) / lin(2) = 0.7213475204, so,

f(x)=a+bln(x+1) f(x)=30+0.72[ln(x+1)]

- Admin
- #34

\(\displaystyle f(x)=30+\frac{\ln(x+1)}{2\ln(2)}\)

This would allow you to use the change of base formula to write:

\(\displaystyle f(x)=30+\log_4(x+1)\)

You probably want to stick with the first form though for using the computer to generate a graph.

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- #35

- Aug 13, 2013

- 45

How would would i put my answers into WA?

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- #36

Use the command:

How would would i put my answers into WA?

y=x^2/2+20,y=30+ln(x+1)/(2ln(2)) where x=0 to 8

- Thread starter
- #37

- Aug 13, 2013

- 45

they're both supposed to go up by 50 cents, same amount.

im confused

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- #38