The derivative & tangent line equations | Derivatives introduction | AP Calculus AB | Khan Academy

We're told that the tangent line to the graph of function at the point (2, 3) passes through the point (7, 6). Find f prime of 2.

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 the x-axis, that is the y-axis. Let's see the relevant points here: (2, 3) and (7, 6). So let me go 1, 2, 3, 4, 5, 6, 7 along the x-axis, and I'm going to go 1, 2, 3, 4, 5, and 6 along the y-axis.

Now this point, so we have the point (2, 3). So let me mark that. So (2, 3) is right over there. So that's (2, 3), and we also have the point (7, 6). (7, 6) is going to be right over there.

Now let's remind ourselves what they're saying. They say the tangent line to the graph of function f at this point passes through the point (7, 6). So if it's the tangent line to the graph at that point, it must go through (2, 3). That's the only place where it intersects our graph, and it goes through (7, 6).

You 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 to f right at (2, 3), and it goes through the point (7, 6).

And so we don't know anything other than f, but we can imagine what f looks like. Our function f 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 2, they're really saying what is the slope of the tangent line when x is equal to 2. So when x is equal to 2, well, the slope of the tangent line is the slope of this line. 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 this derivative.

It's going to be the slope of the tangent line because this is the tangent line. So let's do that! As we know, slope is change in y over change in x.

So if we change our y to go from (2, 3) to (7, 6), our change in x... change in x; we go from x equals 2 to x equals 7, so our change in x is equal to 5. And our change in y, our change in y; we go from y equals 3 to y equals 6, so our change in y is equal to 3.

So our change in y over change in x is going to be 3 over 5, which is the slope of this line, which is the derivative of the function at 2 because this is the tangent line at x equals 2.

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 1?

All right, 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 (-1, 3) 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 (-1, 3); it's going to be on our function. And we also know that g prime of negative 1 is equal to negative 2. So the slope of the tangent line right at that point on our function is going to be negative 2.

That's what that tells us—the slope of the tangent line when x is equal to negative 1 is equal to negative 2. 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 2 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 has 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.

And we can make g do all sorts of crazy things after that, but all we really care about is the 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 2. So we could say y is equal to negative 2x plus b. And then to solve for b, we know that the point (-1, 3) is on this line. And this goes back to some of your Algebra 1 that you might have learned a few years ago.

So let's substitute -1 and 3 for x and y. So when y is equal to 3, we have 3 equals negative 2 times negative 1 plus b. And so let's see, this is negative 2 times negative 1, which is positive 2, and so if you subtract 2 from both sides, you get 1 is equal to b.

And there you have it! That is the equation of our line: y is equal to -2x + 1. 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.