WHAT IS A TANGENT LINE

TANGENT LINE EXPLANATION WITH EXAMPLE AND GRAPH

The first problem that we’re going to take a look at is the tangent line problem. Before getting into this problem it would probably be best to define a tangent line.

A tangent line to the function $f\left(x\right)$ at the point $x=a$ is a line that just touches the graph of the function at the point in question and is “parallel” (in some way) to the graph at that point. Take a look at the graph below.

In this graph the line is a tangent line at the indicated point because it just touches the graph at that point and is also “parallel” to the graph at that point. Likewise, at the second point shown, the line does just touch the graph at that point, but it is not “parallel” to the graph at that point and so it’s not a tangent line to the graph at that point.

At the second point shown (the point where the line isn’t a tangent line) we will sometimes call the line a secant line.

NOW WE EXPLAIN IT WITH THE HELP OF A EXAMPLE

Example 1 Find the tangent line to $f\left(x\right)=15-2x^2$ at $x=1$

Solution

We know from algebra that to find the equation of a line we need either two points on the line or a single point on the line and the slope of the line. Since we know that we are after a tangent line we do have a point that is on the line. The tangent line and the graph of the function must touch at $x$ = 1 so the point $\left(1,f\left(1\right)\right)=\left(1,13\right)$ must be on the line.
Now we reach the problem. This is all that we know about the tangent line. In order to find the tangent line we need either a second point or the slope of the tangent line. Since the only reason for needing a second point is to allow us to find the slope of the tangent line let’s just concentrate on seeing if we can determine the slope of the tangent line.
At this point in time all that we’re going to be able to do is to get an estimate for the slope of the tangent line, but if we do it correctly we should be able to get an estimate that is in fact the actual slope of the tangent line. We’ll do this by starting with the point that we’re after, let’s call it $P=\left(1,13\right)$. We will then pick another point that lies on the graph of the function, let’s call that point $Q=\left(x,f\left(x\right)\right)$.
For the sake of argument let’s take choose $x=2$ and so the second point will be $Q=\left(2,7\right)$. Below is a graph of the function, the tangent line and the secant line that connects $P$ and $Q$.

see below in graph

We can see from this graph that the secant and tangent lines are somewhat similar and so the slope of the secant line should be somewhat close to the actual slope of the tangent line. So, as an estimate of the slope of the tangent line we can use the slope of the secant line, let’s call it $m_{PQ}$, which is,
$$m_{PQ}=\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{7-13}{1}=-6$$
Now, if we weren’t too interested in accuracy we could say this is good enough and use this as an estimate of the slope of the tangent line. However, we would like an estimate that is at least somewhat close the actual value. So, to get a better estimate we can take an $x$ that is closer to $x=1$ and redo the work above to get a new estimate on the slope. We could then take a third value of $x$ even closer yet and get an even better estimate.
In other words, as we take $Q$ closer and closer to $P$ the slope of the secant line connecting $Q$and $P$ should be getting closer and closer to the slope of the tangent line. If you are viewing this on the web, the image below shows this process.

As you can see we moved $Q$ in closer and closer to $P$the secant lines does start to look more and more like the tangent line and so the approximate slopes (i.e. the slopes of the secant lines) are getting closer and closer to the exact slope. Also, do not worry about how I got the exact or approximate slopes. We’ll be computing the approximate slopes shortly and we’ll be able to compute the exact slope in a few sections.
In this figure we only looked at $Q$’s that were to the right of $P$, but we could have just as easily used $Q$’s that were to the left of $P$ and we would have received the same results. In fact, we should always take a look at $Q$’s that are on both sides of $P$. In this case the same thing is happening on both sides of $P$. However, we will eventually see that doesn’t have to happen. Therefore, we should always take a look at what is happening on both sides of the point in question when doing this kind of process.
So, let’s see if we can come up with the approximate slopes we showed above, and hence an estimation of the slope of the tangent line. In order to simplify the process a little let’s get a formula for the slope of the line between $P$ and $Q$, $m_{PQ}$, that will work for any $x$ that we choose to work with. We can get a formula by finding the slope between $P$ and $Q$using the “general” form of $Q=\left(x,f\left(x\right)\right)$.
$$m_{PQ}=\frac{f\left(x\right)-f\left(1\right)}{x-1}=\frac{15-2x^2-13}{x-1}=\frac{2-2x^2}{x-1}$$
Now, let’s pick some values of $x$ getting closer and closer to $x=1$, plug in and get some slopes.

$x$	$m_{PQ}$	$x$	$m_{PQ}$
2	-6	0	-2
1.5	-5	0.5	-3
1.1	-4.2	0.9	-3.8
1.01	-4.02	0.99	-3.98
1.001	-4.002	0.999	-3.998
1.0001	-4.0002	0.9999	-3.9998

So, if we take $x$’s to the right of 1 and move them in very close to 1 it appears that the slope of the secant lines appears to be approaching -4. Likewise, if we take $x$’s to the left of 1 and move them in very close to 1 the slope of the secant lines again appears to be approaching -4.
Based on this evidence it seems that the slopes of the secant lines are approaching -4 as we move in towards $x=1$, so we will estimate that the slope of the tangent line is also -4. As noted above, this is the correct value and we will be able to prove this eventually.
Now, the equation of the line that goes through$$\left(a,f\left(a\right)\right)$$is given by
$$y=f\left(a\right)+m\left(x-a\right)$$
Therefore, the equation of the tangent line to $f\left(x\right)=15-2x^2$ at $x=1$ is
$$y=13-4\left(x-1\right)=-4x+17$$

There are a couple of important points to note about our work above. First, we looked at points that were on both sides of $x=1$. In this kind of process it is important to never assume that what is happening on one side of a point will also be happening on the other side as well. We should always look at what is happening on both sides of the point. In this example we could sketch a graph and from that guess that what is happening on one side will also be happening on the other, but we will usually not have the graphs in front of us or be able to easily get them.
Next, notice that when we say we’re going to move in close to the point in question we do mean that we’re going to move in very close and we also used more than just a couple of points. We should never try to determine a trend based on a couple of points that aren’t really all that close to the point in question.
The next thing to notice is really a warning more than anything. The values of $m_{PQ}$ in this example were fairly “nice” and it was pretty clear what value they were approaching after a couple of computations. In most cases this will not be the case. Most values will be far “messier” and you’ll often need quite a few computations to be able to get an estimate. You should always use at least four points, on each side to get the estimate. Two points is never sufficient to get a good estimate and three points will also often not be sufficient to get a good estimate. Generally, you keeping picking points closer and closer to the point you are looking at until the change in the value between two successive points is getting very small.
Last, we were after something that was happening at $x=1$ and we couldn’t actually plug $x=1$ into our formula for the slope. Despite this limitation we were able to determine some information about what was happening at $x=1$ simply by looking at what was happening around $x=1$. This is more important than you might at first realize and we will be discussing this point in detail in later sections.
Before moving on let’s do a quick review of just what we did in the above example. We wanted the tangent line to $f\left(x\right)$ at a point $x=a$. First, we know that the point $P=\left(a,f\left(a\right)\right)$ will be on the tangent line. Next, we’ll take a second point that is on the graph of the function, call it $Q=\left(x,f\left(x\right)\right)$ and compute the slope of the line connecting $P$ and $Q$ as follows,
$$m_{PQ}=\frac{f\left(x\right)-f\left(a\right)}{x-a}$$
We then take values of $x$ that get closer and closer to $x=a$ (making sure to look at $x$’s on both sides of $x=a$ and use this list of values to estimate the slope of the tangent line, $m$.
The tangent line will then be,
$$y=f\left(a\right)+m\left(x-a\right)$$