Are my answers correct? Quadratic and logarithmic modeling

[needalgebra]

Thanks!

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!

 

[MarkFL]

Re: Are my answer correct?

For the quadratic model, you want:

$$f(x)=ax^2+b$$

where:

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

$$f(1)=a(1)^2+b=a+b=20.5$$

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:

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

where:

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

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

I would use substitution here as well.

 

[needalgebra]

Answers:

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)]

 

[MarkFL]

Looks good, although I would choose to express the parameter $b$ for the logarithmic model in exact form:

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

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

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

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

 

[needalgebra]

the graph was going to be my next question! - sorry for being a pain in the butt. :p

How would would i put my answers into WA?

 

[MarkFL]

the graph was going to be my next question! - sorry for being a pain in the butt. :p

How would would i put my answers into WA?

Use the command:

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

 

[needalgebra]

why is logarithmic model pretty much staying the same...

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

im confused

 

[MarkFL]

The both go up by 50 cents for the first gigabyte, but only a linear function will change at a constant rate.