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I was given the point P (4,2) and Q $(x,\sqrt{x})$ and was told to find the secant line. I forgot what the difference between secant and tangent was but I thought it might have something to do with that formula I was suppose to memorize but forgot. As far as I could make sense of this the formula should be something like $\frac{(a-h)+a}{h}$ so I used that and I got 1 which I think is wrong so I wrote 2. I was then given points like 3.5 3.9 3.99 and such and told to find the tangent lines I think and then estimate the tangent I think. I didn't know how to do this because I got -1 and 1 which can't be right so I made up numbers getting closer to two. I was then told to estimate the secant lines and I guessed 2 because I don't know how to do that.

I tried to take the derivative of it which gave me $\frac{1}{2}x^\frac{-3}{4}$ which gave me .56 something or so so I just made up an answer and guessed 2 because 4/2=2.

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Do you have some sort of notes with definitions you could consult? –  anon Jan 19 '12 at 2:11
Someone has downvoted the question and both the answers. @downvoter: Care to explain the vote? –  Srivatsan Jan 19 '12 at 4:52
@Jordan: That is not a fair assumption, nor respectful to the people who are trying to help you with your question. Please base your voting decisions on the content of the posts. Upvote good and helpful posts, and please only downvote low quality posts. –  Jonas Meyer Jan 19 '12 at 5:09
@Jordan: Your frustration is understandable. In particular, I don't know why no one has accompanied a downvote to your question with a comment, for example suggesting improvement. I just mean that (1) It seems very unlikely that the people who are trying to help you are also downvoting your question, and in any case should not be assumed, and (2) We want to keep the voting system based on the content, not on speculations of the poster's other activity. One further note: (3) If you vote in retaliation like this, it might make users less likely to want to comment to explain their vote. –  Jonas Meyer Jan 19 '12 at 5:19
I didn't vote either way on the question, though I was sorely tempted to -1 it (but won't) after I learned mine was downvoted for no good reason. I'm sorry about your issues with college, and I'm far more concerned about your well-being than petty votes on MSE, but the only thing I can really do to help out over the internet is write whatever comes to mind as an answer here and let things take their course afterwards. Live and learn, Jordan. :) –  anon Jan 20 '12 at 7:03

2 Answers 2

Secant lines go through two points, while tangent lines meet smoothly with the curve at one point:

$\hskip 2in$ pic

$\hskip 1.5in$ pic2

A secant line for two points $(x_A,y_A)$ and $(x_B,y_B)$ is given by a rise-over-run formula:

$$\frac{y-y_A}{x-x_A}=\frac{y_B-y_A}{x_B-x_A}. \tag{S}$$

This of course might look awfully daunting to remember. How to memorize it? The key is to see what both sides of the equation stand for: the left side is the slope $m$ (rise/run) of the line between the points $(x,y)$ and $(x_A,y_A)$ while the right side is the slope between $(x_A,y_A)$ and $(x_B,y_B)$. So if you want, you can use this diagram I made up as a mnemonic to be unpackaged whenever:

$\hskip 1.75in$ pic3

(This says the slope of $\overline{AB}$ equals that of $\overline{BC}$. Here we relabel $(x,y)$ as $C$.) This of course makes sense: any point $(x,y)$ that is off this line will not form a line with $A$ having the same slope as $\overline{AB}$.

Now for a curve given by $(x,f(x))$ in coordinates (i.e. $y=f(x)$), secant lines will indeed have slope given by $$m=\frac{f(b)-f(a)}{b-a},$$ which is similar to what you remember if we let $x=a$ and $b=x+h$.

Tangent lines are trickier. We must let the two points $A$ and $B$ approach each other indefinitely, and in the limit we get a derivative in the expression:


These two formulas for secant and tangent lines, $(S)$ and $(T)$ above, are what you need to (1) remember consistently and (2) know how to apply in actual problems.

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I thought I understood how these worked but this thoroughly confused me. I have no idea what is wrong with me but I don't think I am smart or hard working enough for this class after taking it a second time. –  user138246 Jan 19 '12 at 2:46
@Jordan, please try to only add comments on the topic. –  Mariano Suárez-Alvarez Jan 19 '12 at 6:18
I don't understand this at all, how does that formula work? Why are the secant lines equal with slope? Why can't they cross the same line twice and have different slopes? I can think of many examples where that works. –  user138246 Jan 20 '12 at 16:17

It's a trick question, sort of. A secant line is any line that touches the graph in (at least) two places; the tangent line is what you get when you let the two places get infinitely close together. The secant line between $P$ and $Q$ is just the line between those two points; for points $(a,b)$ and $(c,d)$, the line will have the expression

$$y = \frac{d-b}{c-a}(x-a) + b$$

For the specific points they give you, the line will be a function of, let's say, $s$, where $Q$ is the point $(s,\sqrt{s})$ (I would use $x$ to match your notation, but then it conflicts with the convention of using $x$ and $y$ for a line.)

For the rest of your question, are you sure you're remembering the problem correctly? I think they might be asking you to use the secant lines to estimate the tangent line (to the function $f(x) = \sqrt{x}$) at the point $(4,2)$.

You can do this by noting that the slop of the secant line between $P(4,2)$ and $Q(x,\sqrt{x})$ is $(\sqrt{x}-2)/(x-4)$ and then plugging in the numbers they gave you for $x$; since the $x$-values are increasingly close to $4$, the slopes of the secants will be close to the slope of the tangent line.

Taking the derivative is a good way to check your work, since you then have the exact value you're supposed to be approximating. In this case, $\frac{d}{dx}\sqrt{x}$ is $$\frac{d}{dx}x^{1/2}=\frac{1}{2}x^{1/2-1}=\frac{1}{2\sqrt{x}}$$ so that the derivative at $x=4$ is $1/4$.

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