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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 2

Lesson 1: Defining average and instantaneous rates of change at a point- Newton, Leibniz, and Usain Bolt
- Derivative as a concept
- Secant lines & average rate of change
- Secant lines & average rate of change
- Derivative notation review
- Derivative as slope of curve
- Derivative as slope of curve
- The derivative & tangent line equations
- The derivative & tangent line equations

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# The derivative & tangent line equations

Discover how the derivative of a function reveals the slope of the tangent line at any point on the graph. We'll explore how to use this powerful tool to determine the equation of the tangent line, enhancing our understanding of instantaneous rates of change.

## Want to join the conversation?

- I thought a tangent line could only touch one point on the graph?(37 votes)
- Technically, a tangent line is one that touches a curve at a point without crossing over it. Essentially, its slope matches the slope of the curve at the point. It does not mean that it touches the graph at only one point. It is, in fact, very easy to come up with tangent lines to various curves that intersect the curve at other points.

The key here is that the tangent should only touch the curve at the point of interest.(83 votes)

`g'(-1)=-2`

-- doesn't that mean that the tangent goes through (-1,-2)? Sal's tangent doesn't.(13 votes)- I was confused by this as well. I was thinking, incorrectly, that "g'(-1) = -2" means that the tangent line must go through (-1, -2), as you said. But the derivative, g', doesn't tell us what points the tangent line goes through. It tells us only the
*slope*of the line for a given x. The value of the*slope*of the tangent line could be 50 billion, but that doesn't mean that the tangent line goes through 50 billion. In fact, the tangent line must go through the point in the original function, or else it wouldn't be a tangent line. The derivative function, g', does go through (-1, -2), but the tangent line does not. It might help to think of the derivative function as being on a second graph, and on the second graph we have (-1, -2) that describes the tangent line on the first graph:*at x = -1 in the first graph, the slope is -2*.(39 votes)

- I might be missing something, but why did we choose to represent the line with equation y=mx+b ?

Could we use any other equation and still get the same result for value of 'b' ?(7 votes)- He chose to use y=mx+b because a tangent line, or the derivative of a function will always be a straight line, and that equation (y=mx+b) is how we show the line. The 'b' value is just the y-intercept. It is where the line hits the y/vertical axis. But I'm just starting to study this stuff so don't rely on my answer too much lol. Good luck.(16 votes)

- I ran into this question on a quiz:

The tangent line to the graph of function g at the point (-6, -2), passes through the point (0,2).

Find g'(-6).

The correct solution is:

((-2)-2) / (-6-0) = -4 / -6 =**2/3**

But could it also be this?

(2-(-2)) / 0-(-6) = 0 / -6 =**0**

Thanks!(0 votes)- Your setup in the second version is correct, but 2-(-2)=4, and 0-(-6)=6, so the slope is still 4/6=2/3.(13 votes)

- I'm happy with the y = mx + c form, but in the questions that follow this section, they were of the form (y + a) = b(x + d).

It wasn't obvious to me how to convert one to the other.(2 votes)- That seems to be the point-slope form. It's used to get the equation when a point and the slope are given. This is actually derived from the slope formula itself.

Let the slope = m and let point (p,q) exist on the line. We need to find the equation of the line. Let's take an arbitrary point (x,y). So, from the definition of a slope.

m = (y-q)/(x-p)

Rearranging, you get (y-q) = b(x-p)

And that's exactly what you wrote, just with different variables.

Honestly speaking, such questions can almost always be done with the definition of a slope. So, you don't really need to bother learning all the different formulae. They're just iterations of the same thing.(8 votes)

- consider a function f(x)=x^2 its derivative will be f'(x)=2x and f'(2) will be 4

am i right?(5 votes) - Couldn't you have also found the y-intercept of the tangent line just by looking at, y'know, where it intercepts the y-axis? I knew from looking at the line that b = 1 because it's very obvious that the line intercepted the y-axis at y = 1. Did he do the algebra just for the sake of clarification or have I been finding y-intercepts wrong my whole life? lol(3 votes)
- A lot of times it won't be so easy to tell where the y intercept is. Basically when the y intercept is not exactly on a marked number.(3 votes)

- Isn't "b" in y=mx+b where the function crosses the Y-axis?(2 votes)
- Yes it is. The tangent line is the straight green line he drew, which infact crosses the y-axis at y=1(3 votes)

- please solve this for me and explain it??

Find the value of a, b, c for which the parabola ax^2+bx+c passes through (1, 1) and has a tangent with gradient 7, at point (3, 3).

this formula might be helpful:

y- f(a) = f'(a)(x-a)(0 votes)- Plugging in your point (1, 1) tells us that a+b+c=1.

You also say it touches the point (3, 3), which tells us 9a+3b+c=3. Subtract the first from the second to obtain 8a+2b=2, or 4a+b=1.

The derivative of your parabola is 2ax+b. When x=3, this expression is 7, since the derivative gives the slope of the tangent. So 6a+b=7.

So we have

6a+b=7

4a+b=1

Subtract the second equation from the first to get 2a=6, or a=3. Substitute back in to get 12+b=1, or b= -11.

Since a+b+c=1, this gives us 3-11+c=1

-8+c=1

c=9

So your parabola is 3x²-11x+9(7 votes)

- at5:32how did sal draw the tangent by looking at g'(-1)= -2 ?(2 votes)
- The slope of the tangent line is -2, that is, for every +1 unit of x, y goes down 2 units.(1 vote)

## Video transcript

- [Voiceover] We're told that
the tangent line to the graph of function at the the
point two comma three passes through the point seven comma six. Find f prime of two. So whenever you see something like this, it doesn't hurt to try to visualize it. You might want to draw it
out or just visualize it in your head but since you can't get in my head, I will draw it out. So let me draw the information
that they are giving us. So that's x axis and that is the y axis. Let's see the relevant points
here at two comma three and seven comma six. So let me go, one, two, three, four, five, six, seven, along the x axis and I'm going to go one, two, three, four, five, and six, along the y axis. And now this point, so we have the point two comma three, so let me mark that, so two comma three is right over there, so that's two comma three and we also have the point seven comma six. Seven comma six is going
to be right over there. Seven comma six. Let us remind ourselves
what they're saying. They're saying the tangent
line to the graph of function f at this point passes through the point seven comma six. So if it's the tangent line
to the graph at that point, it must go through two comma three, that's the only place where
it intersects our graph and it goes through seven comma six. We only need two points to define a line and so the tangent line
is going to look like, it's going to look like, let me see if I can, no that's not right. Let me draw it like it's going to look. Oh that's not exactly right. Let me try one more time. Okay, there you go. So the tangent line is
going to look like that. It goes, it's tangent two
f right at two comma three and it goes through the
point seven comma six and so we don't know anything other than f but we can imagine what f looks like. Our function f could, so our function f, it could look something like this. It just has to be tangent so that line has to be
tangent to our function right at that point. So our function f could
look something like that. So when they say, find f prime of two, they're really saying, what is the slope of the tangent line when x is equal to two? So when x is equal to two, well the slope of the tangent line is the slope of this line. They gave us, they gave
us the two points that sit on the tangent line. So we just have to figure out its slope because that is going to
be the rate of change of that function right over there, its derivative. It's going to be the
slope of the tangent line because this is the tangent line. So let's do that. So as we know, slope is change in y over change in x. So if we change our, to go from two comma
three to seven comma six, our change in x, change in x, we go from x equals two to x equals 7 so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six. So our change in y is equal to three. So our change in y over change in x is going to be three over five which is the slope of this line, which is the derivative
of the function at two because this is the tangent
line at x equals two. Let's do another one of these. For a function g, we are given that g of
negative one equals three and g prime of negative one
is equal to negative two. What is the equation of the
tangent line to the graph of g at x equals negative one? Alright, so once again I think it will be helpful to graph this. So we have our y axis, we have our x axis and let's see. We say for function g we are
given that g of negative one is equal to three. So the point negative one
comma three is on our function. So this is negative one and then we have, one, two, and three. So that's that right over there. That is the point. That is the point
negative one comma three, it's going to be on our function. And we also know that
g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us. The slope of the tangent line, when x is equal to negative
one is equal to negative two. So I could use that information to actually draw the tangent line. So let me see if I can, let me see if I can do this. So it will look, so I think it will, let me just draw it like this. So it's going to go, so it's a slope of negative two is going to look something like that. So as we can see if we move
positive one in the x direction, we go down two in the y direction. So that is a slope of negative two. And so you might say well, where is g? Well we could draw what g could look like. G might look something like this. Might look something like
that right over there where that is the tangent line or you can make g do all sorts
of crazy things after that but all we really care about is equation for this green line. And there's a couple of
ways that you could do this. You could say, well look
a line is generally, there's a bunch of different
ways where you can define the equation for a line. You could say a line has a form y is equal to mx plus b where m is the slope and
b is the y intercept. Well we already know what
the slope of this line is. It is negative two. So we could say y is
equal to negative two. Negative two times x. Times x plus b. And then to solve for b, we know that the point
negative one comma three is on this line and this goes back to
some of your Algebra I that you might have
learned a few years ago. So let's substitute negative
one and three for x and y. So when y is equal to three, so three, three is equal to, is equal to negative two, negative two times x. Times negative one, times negative one plus b. Plus b. And so let's see, this is negative two times
negative one is positive two. And so if you subtract
two from both sides, you get one is equal to b. And there you have it. That is equation of our line. Y is equal to negative two x plus one. And there's other ways that
you could have done this, you could have written the
line in point-slope form or you could have done it this way. You could have written it in standard form but at least this is the way
my brain likes to process it.