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) = x2. 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) = x2 just to see what it looks like.

Graph: f(x) = x2

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) = x2 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 (1,12) or (1,1). The point-slope form of the equation of a line is y - y1 = m(x - x1). 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) = x2 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) = x2, 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) = n2. So our point would be (n,n2). Now we can plug into the point-slope form of the equation of a line to find our arbitrary tangent line to f(x) = x2. We have m = 2n, x1 = n, and y1 = n2. Plugging into the equation y - y1 = m(x - x1) we get y - n2 = 2n(x - n). Simplifying gives us y = (2n)(x - n) + n2 and then y = (2n)x - n2. We can now graph this and use Graphing Calculator to vary the n value.

Graph: f(x) = x2 and y = (2n)x - n2, value of n varies

Conclusions:

We have shown how a tangent line can be found at any point along the graph f(x) = x2. 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: x1 = n, y1 = 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).