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The point $P(4, 24)$ lies on the curve $y = x^2 + x + 4$. If $Q$ is the point $(x, x^2 + x + 4 )$, find the slope of the secant line $PQ$ for the following values of $x$.

If $x= 4.1$, the slope of $PQ$ is:

and if $x= 4.01$, the slope of $PQ$ is:

and if $x= 3.9$, the slope of $PQ$ is:

and if $x= 3.99$, the slope of $PQ$ is:

Based on the above results, guess the slope of the tangent line to the curve at $P(4, 24)$.

For this problem, should I just plug in the x values given into the y equation? Then the slope would be...??

HELP! Please!

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Would you happen to remember how to get the slope of a segment joining two points? – J. M. Oct 8 '11 at 5:09
You have the point $P(4,24)$ on the curve. If $x=4.1$, what is the corresponding $y$ on the curve? Yes, plug in, you get $Q(4.1,?)$. The problem asks you then to find the slope of the line $PQ$, so you are not quite through. – André Nicolas Oct 8 '11 at 5:22
What happened to your other, quite related, question? You were asked two questions in the comments there, which were meant to guide you towards a solution, but you answered none. You were also given two answers there, but you reacted to none, commented none and accepted none. Well, well, well... – Did Oct 8 '11 at 18:38

Since you will use several values of $x$, it is probably easiest to find a function for the slope, in terms of $x$, and then plug in the various values for $x$. That is, the slope of the secant line $PQ$ is the rise over run (change in $y$ over change in $x$):

$$m(x) = \frac{x^2 + x + 4 - 24}{x - 4}$$

So, $m(x)$ gives the slope for any particular value of $x$. A practical reason to do this is, for example on a TI-83 or TI-84 or something like it, you can now type in that function to $Y_1$ and then go to the Table and plug in the various $x$ values and you get the slopes immediately. And, since it is so fast, you can check this for more values than the question even asks for to get an even better intuition. Or, you could use Wolfram Alpha to accomplish the same thing. For example, type in:

Evaluate (x^2 + x + 4 - 24)/(x - 4) at x = 4.1, 4.01, 4.001, 3.9, 3.99, 3.999

Back to the problem,

$\begin{align*} m(4.1) =& \frac{4.1^2 + 4.1 + 4 - 24}{4.1 - 4} = \frac{0.91}{0.1} = 9.1 \\ m(4.01) =& \frac{4.01^2 + 4.01 + 4 - 24}{4.01 - 4} = \frac{0.0901}{0.01} = 9.01 \\ m(4.001) =& \frac{4.001^2 + 4.001 + 4 - 24}{4.001 - 4} = \frac{0.009001}{0.001} = 9.001 \\ m(3.9) =& = \frac{3.9^2 + 3.9 + 4 - 24}{3.9 - 4} = \frac{-0.89}{-0.1} = 8.9 \\ m(3.99) =& = \frac{3.99^2 + 3.99 + 4 - 24}{3.99 - 4} = \frac{-0.0899}{-0.01} = 8.99 \\ m(3.999) =& = \frac{3.999^2 + 3.999 + 4 - 24}{3.999 - 4} = \frac{-0.008999}{-0.001} = 8.999 \end{align*}$

From this, we would probably guess that the slope of the tangent line when $x = 4$ is 9. This does not guarantee that we are right, but assuming the function is reasonably well behaved, we could be pretty confident in this guess.

And, once you learn how to calculate derivatives, you will find this is correct as the slope of the tangent line at any $x$ value is the derivative. Since

$$y'(x) = 2x + 1$$

we see that

$$y'(4) = 9$$

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