Consider the function y = 1/3 x2 - 4 which is graphed on the right. This function is a curve, commonly called a parabola. Does it make sense to talk about the slope of a curve like this, when clearly it gets steeper and steeper the further you get from the center? The slope is changing, point by point, and getting bigger!

In fact, you can describe the slope of this curve, if you are willing to work out the slope at some particular point on the curve.

In this exercise, we will show you how to determine the steepness (slope) of the curve at the point (3, -1), shown here by the red dot.

The slope of the curve at (3, -1) is exactly the same as the slope of the line at (3, -1). This line touches the curve only at (3, -1) and nowhere else; it is the tangent to the curve at (3, -1).

Knowing the slope of this line, we would also know the slope of the curve at that point.

Unfortunately, the point (3, -1) is the only point we know on this line, and we can't determine slope from just one point; we need two.

But if there was some way to figure out the slope of that line, we would then have the slope of the curve at (3, -1). What we're going to do is show you how to determine the slope of that line ... in a rather round-about way!

The method we're going to use is as follows. We're going to find the slope of some lines that pass through (3, -1) by picking various second points. We'll pick those points such that the lines we sketch get closer and closer to the line we want. By looking at what the values for the slopes of these lines seem to be approaching, we'll be able to guess the slope of the tangent line at (3, -1).

Let's begin by finding another point on the curve.

Pick x = -1 at random. The corresponding y value is -3.7 (from the equation of the curve y = 1/3 x2 - 4).

Join this point to the point (3, -1) to make a line. This line cuts through the curve in two places, and is called a secant.

We can work out its slope, because we know two points on this line, the points:

(3, -1)    and   (-1, -3.7)

Here is the slope calculation:


The slope of this secant to the curve is about 0.68

That's not the line we're interested in, but at least we know the slope of that tangent will be bigger than  0.68

Now lets move closer to  (3, -1) and try again.



Moving to the right, another point on the curve at x = 0 is the point  (0, -4) ... again found using the parabola's equation.

Join this point to  (3, -1) and make a new secant.

This line is getting a little closer to the line we want; we can find its slope like we did for the previous secant; using the points  (0, -4)  and  (3, -1),  the slope works out to be 1

Clearly the secants are getting steeper.

Let's do it again ...



Move closer; let's try x = 1. We get the point  (1, -3.7), again obtained from the table of values for the equation of the curve.

Drawing the secant to (3, -1) and working out its slope gives the value 1.35

This secant is even steeper.

But we're not there yet ... lets find a point even closer to  (3, -1)




Using the point (2, -2.7) and the point  (3, -1), the secant this time is steeper still; in fact, its slope turns out to be 1.7

Can you see where we're heading? The aim is to find the slopes of secants that are getting closer and closer to the tangent at (3, -1).

Let's do another one ...



This time we've moved very close to (3, -1)

We're at the point (2.5, -1.9)

This secant has slope 1.8

This is very close to the slope of the tangent right at the point (3, -1). Let's try one more closer point ...

Using the point (2.8, -1.39), which is right next to our point  (3, -1), we get a secant line with slope 1.95

This should be almost identical to the slope of the tangent exactly at (3, -1)

We could try closer points, but you should be able to guess what the slope of the tangent at (3, -1) will be ...

Here are the slopes of the secants we've found so far, as we got closer and closer to the tangent line:

0.68,  1,  1.35,  1.7,  1.8,  1.95  ...
It looks very much like the slopes are getting closer and closer to 2

We could be sure of this by doing the same exercise all over again, from the other side of (3, -1). If we did, we'd find that the slopes got smaller and smaller as our secant lines got closer and closer to the tangent; the slopes would again seem to approach the value 2



CONCLUSION:

By finding the slopes of secants, as the secants were drawn from points closer and closer to  (3, -1), it was determined that the slope of the tangent at  (3, -1), or the slope of the curve at  (3, -1), was  2

This is the 'method of secants' for finding the slope of a curve at a point, and it works for any curve.

You might want to visit our 'Derivative of a Polynomial Function' page that introduces calculus, and a quicker method to find the slope at a point.



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