# Thread: Slope of a tangent line and weird world problem

1. ## Slope of a tangent line and weird world problem

Three problems:

1) find the slope of the tangent line for the graph f(x)= x^2 at (2,4).

2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)

And

A rancher has 300 feet of fence to enclose two adjacent pastures. (See figure) Express the total area A of the two pastures as a function of x. What is the domain of A?

Graph the area function (I suppose I can do this once I find the equation) and estimate the dimensions that yield the maximum amount of area for the pastures.

Find the dimensions that yield the maximum amount of area for the pastures by completing the square.

THANKS for reading this and trying to help

2. Originally Posted by superfly8912
Three problems:

1) find the slope of the tangent line for the graph f(x)= x^2 at (2,4).
Find $f'(2)$

Spoiler:
$f(x) = x^2 \Rightarrow f'(x) = 2x \Rightarrow f'(2) = 2\times 2 = 4$

3. Originally Posted by superfly8912

2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)
Let the slope of the line be m.

Find $m = \frac{f(2+h)-f(4)}{2+h-2}$

Spoiler:
$m = \frac{f(2+h)-f(4)}{2+h-2} = \frac{(2+h)^2-(4)^2}{h} =\frac{4+4h+h^2-16}{h} =\frac{4h+h^2-12}{h}$

Originally Posted by superfly8912

A rancher has 300 feet of fence to enclose two adjacent pastures. (See figure) Express the total area A of the two pastures as a function of x. What is the domain of A?

Graph the area function (I suppose I can do this once I find the equation) and estimate the dimensions that yield the maximum amount of area for the pastures.

Find the dimensions that yield the maximum amount of area for the pastures by completing the square.
Consider the perimeter of the fence.

$x+x+x+y+y+y+y = 300$

$3x+4y = 300$

Rearrange for y

$y = \frac{300-3x}{4}$

Now Area $= 2xy$ as there are 2 pastures

$A= 2xy = 2x\frac{300-3x}{4} =\frac{600x-6x^2}{4}$

Now graph this function. It is a quadratic so you can find the turning point (or the maximum) using symmetry.

4. Thank you for helping me out but I'm having trouble. Why did you square the numerator in number two? I happen to have the answer for this problem and apparently its 4 + h (but then again it could be a typo)...

5. Originally Posted by superfly8912

2) Find the slope of the line joining (2,4) and (2+h, f(2+h)) in terms of the nonzero number h. (This is also for the graph f(x)= x^2)
$f(x)= x^2 \Rightarrow f(2+h) = (2+h)^2 = (2+h)(2+h) = 4+4h+h^2$

6. Originally Posted by pickslides
Let the slope of the line be m.

Find $m = \frac{f(2+h)-f(4)}{2+h-2}$
Should be:
$m = \frac{f(2+h)-f(2)}{2+h-2} = \frac{(2+h)^2 - (2)^2}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$

7. Originally Posted by Defunkt
Should be:
$m = \frac{f(2+h)-f(2)}{2+h-2} = \frac{(2+h)^2 - (2)^2}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$
Yes I see... Nice work Defunkt!

Or even easier again

$m = \frac{f(2+h)-4}{2+h-2} = \frac{(2+h)^2 - 4}{h} = \frac{h^2 + 4h + 4 - 4}{h} = h+4$