Welcome to our community

Be a part of something great, join today!

Are my answers correct? Quadratic and logarithmic modeling

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You don't seem to quite have the correct equation. To solve this would require using a numeric root finding technique. By the graph it looks like $x$ is closer to 42.
 

needalgebra

Member
Aug 13, 2013
45
i have no diea how to do that, do you mind showing me how?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For the functions involved, Newton's method would be extremely tedious, and unless you understand differential calculus it would not be worth my effort to crank it out manually. For some reason, W|A will only give the root in which we are not interested.

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

needalgebra

Member
Aug 13, 2013
45
I got this....

36.9999 = y
42 = x
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle (x,y)\approx(42,37)\)

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

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

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

needalgebra

Member
Aug 13, 2013
45
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

Administrator
Staff member
Feb 24, 2012
13,775
Re: Are my answer correct?

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.
 

needalgebra

Member
Aug 13, 2013
45
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

Administrator
Staff member
Feb 24, 2012
13,775
Looks good, although I would choose to express the parameter $b$ for the logarithmic model in exact form:

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

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.
 

needalgebra

Member
Aug 13, 2013
45
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Aug 13, 2013
45
why is logarithmic model pretty much staying the same...

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

im confused
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The both go up by 50 cents for the first gigabyte, but only a linear function will change at a constant rate.