__Assignment 2 Write-Up__
**The Problem:**

Explore the relationship between the graph of a function and the graph of the derivative
of the function. Given a function f(x), what is the equation of the tangent line at any given point. Using
Graphing Calculator 2.2, graph f(x) and the tangent line.

**The Strategy:**

We are going to explore the function **f(x) = x**^{2}. We know quite a bit about the derivative
of a function from Calculus, so we'll begin there and progress to a more general case.

**Foundation:**

Let's begin by graphing **f(x) = x**^{2} just to see what it looks like.

__Graph:__ f(x) = x^{2}

We know that the derivative of a function ( f' ) gives the slope of a line tangent to the function at
any given point x. (The slope, then, would be f'(x).) The derivative of **f(x) = x**^{2} is **f'(x) = 2x**.
Therefore, the slope of the tangent line at **x = 1** is **m = 2(1)** or **m = 2**. Since we know the
slope of the tangent line, we can graph the tangent line by using the slope and a point on the tangent line.
The easiest point to use is at the tangent point. Therefore, we will use the point **(1,1**^{2}) or **(1,1)**.
The point-slope form of the equation of a line is **y - y**_{1} = m(x - x_{1}). Plugging into that equation, we get
**y - 1 = 2(x - 1)**. Simplifying, we get **y = x - 1**. So let's graph that line on the same axes as our
original function....

__Graph:__ f(x) = x^{2} and **y = x - 1**

**Exploration:**

Now we need to generalize this. To do this, let's dissect what we did to find the graph of the tangent line
at **x = 1**. We began by finding the derivative of **f(x)** at **x = 1**. Remember that since
**f(x) = x**^{2}, **f'(x) = 2x**. So to take the derivative at any number, n, we would take **f'(n) = 2n**.
This gives us the slope, m, at any point n. Now that we have the slope, we need to find a point on this
arbitrary tangent line. To do that is simple. All we need to do is plug our n directly into the original
function. This gives us **f(n) = n**^{2}. So our point would be **(n,n**^{2}). Now we can plug into the
point-slope form of the equation of a line to find our arbitrary tangent line to **f(x) = x**^{2}. We have
**m = 2n**, **x**_{1} = n, and **y**_{1} = n^{2}. Plugging into the equation **y - y**_{1} = m(x - x_{1}) we get
**y - n**^{2} = 2n(x - n). Simplifying gives us **y = (2n)(x - n) + n**^{2} and then **y = (2n)x - n**^{2}.
We can now graph this and use Graphing Calculator to vary the n value.

__Graph:__ f(x) = x^{2} and **y = (2n)x - n**^{2}, value of n varies

**Conclusions:**

We have shown how a tangent line can be found at any point along the graph **f(x) = x**^{2}. This leads
us to the question: what about finding a tangent line at any point along *any* function? By retracing
our steps from this exploration, it is simple to generalize this activity for any function, f(x). The values
for the point-slope form of the equation are: **x**_{1} = n, **y**_{1} = f(n), **m = f'(n)**. This gives
us the equation: **y - f(n) = f'(n)[x - n]** or **y = f'(n)[x - n] + f(n)**.

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