# 2.1: Revisiting Tangent Lines

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By way of motivation for the definition of the derivative, we return to the discussion of tangent lines that we started in the previous chapter on limits. We consider, in Examples 2.1.2 and 2.1.5, below, the problem of finding the slope of the tangent line to a curve at a point. But let us start by recalling, in Example 2.1.1, what is meant by the slope of a straight line.

##### Example 2.1.1 What is slope.

In this example, we recall what is meant by the slope of the straight line

\begin{align*} y&=\tfrac{1}{2}x+\tfrac{3}{2} \end{align*}

• We claim that if, as we walk along this straight line, our $$x$$–coordinate changes by an amount $$\Delta x\text{,}$$ then our $$y$$–coordinate changes by exactly $$\Delta y = \tfrac{1}{2}\Delta x\text{.}$$
• For example, in the figure on the left below, we move from the point

\begin{gather*} (x_0,y_0)=(1\,,\,2=\tfrac{1}{2}\times 1+\tfrac{3}{2}) \end{gather*}

on the line to the point

\begin{gather*} (x_1,y_1)=(5\,,\,4=\tfrac{1}{2}\times 5+\tfrac{3}{2}) \end{gather*}

on the line. In this move our $$x$$–coordinate changes by

\begin{gather*} \Delta x= 5-1=4 \end{gather*}

and our $$y$$–coordinate changes by

\begin{gather*} \Delta y=4-2=2 \end{gather*}

which is indeed $$\tfrac{1}{2}\times 4=\tfrac{1}{2}\Delta x\text{,}$$ as claimed

• In general, when we move from the point

\begin{align*} (x_0,y_0) &= (x_0, \tfrac{1}{2}x_0+\tfrac{3}{2}) \end{align*}

on the line to the point

\begin{align*} (x_1,y_1) &= (x_1, \tfrac{1}{2}x_1+\tfrac{3}{2}) \end{align*}

on the line, our $$x$$–coordinate changes by

\begin{gather*} \Delta x=x_1-x_0 \end{gather*}

and our $$y$$–coordinate changes by

\begin{align*} \Delta y&=y_1-y_0\\ &=\big[\tfrac{1}{2}x_1+\tfrac{3}{2}\big] -\big[\tfrac{1}{2}x_0+\tfrac{3}{2}\big]\\ &=\tfrac{1}{2}(x_1-x_0) \end{align*}

which is indeed $$\tfrac{1}{2}\Delta x\text{,}$$ as claimed.
• So, for the straight line $$y=\tfrac{1}{2}x+\tfrac{3}{2}\text{,}$$ the ratio $$\tfrac{\Delta y}{\Delta x} =\tfrac{y_1-y_0}{x_1-x_0}$$ always takes the value $$\frac{1}{2}\text{,}$$ regardless of the choice of initial point $$(x_0,y_0)$$ and final point $$(x_1,y_1)\text{.}$$ This constant ratio is the slope of the line $$y=\tfrac{1}{2}x+\tfrac{3}{2}\text{.}$$

Straight lines are special in that for each straight line, there is a fixed number $$m\text{,}$$ called the slope of the straight line, with the property that if you take any two different points, $$(x_0,y_0)$$ and $$(x_1,y_1)\text{,}$$ on the line, the ratio $$\tfrac{\Delta y}{\Delta x}=\tfrac{y_1-y_0}{x_1-x_0}\text{,}$$ which is called the rate of change of $$y$$ per unit rate of change 1 of $$x\text{,}$$ always takes the value $$m\text{.}$$ This is the property that distinguishes lines from other curves.

Other curves do not have this property. In the next two examples we illustrate this point with the parabola $$y=x^2\text{.}$$ Recall that we studied this example back in Section 1.1. In Example 2.1.2 we find the slope of the tangent line to $$y=x^2$$ at a particular point. We generalise this in Example 2.1.5, to show that we can define “the slope of the curve $$y=x^2$$” at an arbitrary point $$x=x_0$$ by considering $$\tfrac{\Delta y}{\Delta x}=\tfrac{y_1-y_0}{x_1-x_0}$$ with $$(x_1,y_1)$$ very close to $$(x_0,y_0)\text{.}$$

##### Example 2.1.2 Slope of secants of $$y=x^2$$.

In this example, let us fix $$(x_0,y_0)$$ to be the point $$(2,4)$$ on the parabola $$y=x^2\text{.}$$ Now let $$(x_1, y_1) = (x_1, x_1^2)$$ be some other point on the parabola; that is, a point with $$x_1 \neq x_0\text{.}$$

• Draw the straight line through $$(x_0,y_0)$$ and $$(x_1,y_1)$$ — this is a secant line and we saw these in Chapter 1 when we discussed tangent lines 2.
• The following table gives the slope, $$\tfrac{y_1-y_0}{x_1-x_0}\text{,}$$ of the secant line through $$(x_0,y_0)=(2,4)$$ and $$(x_1,y_1)\text{,}$$ for various different choices of $$(x_1,y_1=x_1^2)\text{.}$$  $$x_1$$ $$1$$ $$1.5$$ $$1.9$$ $$1.99$$ $$1.999$$ $$\circ$$ $$2.001$$ $$2.01$$ $$2.1$$ $$2.5$$ $$3$$ $$y_1=x_1^2$$ $$1$$ $$2.25$$ $$3.61$$ $$3.9601$$ $$3.9960$$ $$\circ$$ $$4.0040$$ $$4.0401$$ $$4.41$$ $$6.25$$ $$9$$ $$\tfrac{y_1-y_0}{x_1-x_0}$$ $$3$$ $$3.5$$ $$3.9$$ $$3.99$$ $$3.999$$ $$\circ$$ $$4.001$$ $$4.01$$ $$4.1$$ $$4.5$$ $$5$$
• So now we have a big table of numbers — what do we do with them? Well, there are messages we can take away from this table.
• Different choices of $$x_1$$ give different values for the slope, $$\tfrac{y_1-y_0}{x_1-x_0}\text{,}$$ of the secant through $$(x_0,y_0)$$ and $$(x_1,y_1)\text{.}$$ This is illustrated in Figure 2.1.3 below — the slope of the secant through $$(x_0,y_0)$$ and $$(x_1,y_1)$$ is different from the slope of the secant through $$(x_0,y_0)$$ and $$(x'_1,y'_1)\text{.}$$

If the parabola were a straight line this would not be the case — the secant through any two different points on a line is always identical to the line itself and so always has exactly the same slope as the line itself, as is illustrated in Figure 2.1.4 below — the (yellow) secant through $$(x_0,y_0)$$ and $$(x_1,y_1)$$ lies exactly on top of the (red) line $$y=\tfrac{1}{2}x+\tfrac{3}{2}\text{.}$$

• Now look at the columns of the table closer to the middle. As $$x_1$$ gets closer and closer to $$x_0=2\text{,}$$ the slope, $$\tfrac{y_1-y_0}{x_1-x_0}\text{,}$$ of the secant through $$(x_0,y_0)$$ and $$(x_1,y_1)$$ appears to get closer and closer to the value $$4\text{.}$$
##### Example 2.1.5 More on secants of $$y=x^2$$.

It is very easy to generalise what is happening in Example 2.1.2.

• Fix any point $$(x_0,y_0)$$ on the parabola $$y=x^2\text{.}$$ If $$(x_1,y_1)$$ is any other point on the parabola $$y=x^2\text{,}$$ then $$y_1=x_1^2$$ and the slope of the secant through $$(x_0,y_0)$$ and $$(x_1,y_1)$$ is

\begin{align*} \text{slope} &= \frac{y_1-y_0}{x_1-x_0} =\frac{x_1^2-x_0^2}{x_1-x_0} && \text{since }y=x^2\\ & =\frac{(x_1-x_0)(x_1+x_0)}{x_1-x_0} && \text{remember } a^2-b^2 = (a-b)(a+b)\\ & =x_1+x_0 \end{align*}

You should check the values given in the table of Example 2.1.2 above to convince yourself that the slope $$\tfrac{y_1-y_0}{x_1-x_0}$$ of the secant line really is $$x_0+x_1 = 2+x_1$$ (since we set $$x_0=2)\text{.}$$
• Now as we move $$x_1$$ closer and closer to $$x_0\text{,}$$ the slope should move closer and closer to $$2x_0\text{.}$$ Indeed if we compute the limit carefully — we now have the technology to do this — we see that in the limit as $$x_1 \to x_0$$ the slope becomes $$2x_0\text{.}$$ That is

\begin{align*} \lim_{x_1 \to x_0} \frac{y_1-y_0}{x_1-x_0} &= \lim_{x_1 \to x_0} (x_1+x_0) \qquad \text{by the work we did just above}\\ &= 2x_0 \end{align*}

Taking this limit gives us our first derivative. Of course we haven't yet given the definition of a derivative, so we perhaps wouldn't recognise it yet. We rectify this in the next section.

• So it is reasonable to say “as $$x_1$$ approaches $$x_0\text{,}$$ the secant through $$(x_0,y_0)$$ and $$(x_1,y_1)$$ approaches the tangent line to the parabola $$y=x^2$$ at $$(x_0,y_0)$$”. This is what we did back in Section 1.1.

The figure above shows four different secants through $$(x_0,y_0)$$ for the curve $$y=x^2\text{.}$$ The four hollow circles are four different choices of $$(x_1,y_1)\text{.}$$ As $$(x_1,y_1)$$ approaches $$(x_0,y_0)\text{,}$$ the corresponding secant does indeed approach the tangent to $$y=x^2$$ at $$(x_0,y_0)\text{,}$$ which is the heavy (red) straight line in the figure.

Using limits we determined the slope of the tangent line to $$y=x^2$$ at $$x_0$$ to be $$2x_0\text{.}$$ Often we will be a little sloppy with our language and instead say “the slope of the parabola $$y=x^2$$ at $$(x_0,y_0)$$ is $$2x_0$$” — where we really mean the slope of the line tangent to the parabola at $$x_0\text{.}$$

## Exercises

##### Exercise $$\PageIndex{1}$$

Shown below is the graph $$y=f(x)\text{.}$$ If we choose a point $$Q$$ on the graph to the left of the $$y$$-axis, is the slope of the secant line through $$P$$ and $$Q$$ positive or negative? If we choose a point $$Q$$ on the graph to the right of the $$y$$-axis, is the slope of the secant line through $$P$$ and $$Q$$ positive or negative?

##### Exercise $$\PageIndex{2}$$

Shown below is the graph $$y=f(x)\text{.}$$

1. If we want the slope of the secant line through $$P$$ and $$Q$$ to increase, should we slide $$Q$$ closer to $$P\text{,}$$ or further away?
2. Which is larger, the slope of the tangent line at $$P\text{,}$$ or the slope of the secant line through $$P$$ and $$Q\text{?}$$
##### Exercise $$\PageIndex{3}$$

Group the functions below into collections whose secant lines from $$x=-2$$ to $$x=2$$ all have the same slopes.

##### Exercise $$\PageIndex{4}$$

Give your best approximation of the slope of the tangent line to the graph below at the point $$x=5\text{.}$$

##### Exercise $$\PageIndex{5}$$

On the graph below, sketch the tangent line to $$y=f(x)$$ at $$P\text{.}$$ Then, find two points $$Q$$ and $$R$$ on the graph so that the secant line through $$Q$$ and $$R$$ has the same slope as the tangent line at $$P\text{.}$$

##### Exercise $$\PageIndex{6}$$

Mark the points where the curve shown below has a tangent line with slope $$0\text{.}$$

(Later on, we'll learn how these points tell us a lot about the shape of a graph.)

This page titled 2.1: Revisiting Tangent Lines is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Joel Feldman, Andrew Rechnitzer and Elyse Yeager via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.