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#### MarkFL

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

#### MarkFL

Staff member
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.

I got this....

36.9999 = y
42 = x

#### MarkFL

Staff member
$$\displaystyle (x,y)\approx(42,37)$$

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
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.

#### MarkFL

Staff member

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

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

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

How would would i put my answers into WA?

#### MarkFL

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

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

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

im confused