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## Secant lines & average rate of change with arbitrary points

Current time:0:00Total duration:4:19

# Secant line with arbitrary difference

## Video transcript

- [Voiceover] A secant
line intersects the curve y equals the natural log of x at two points, with x-coordinates two and two plus h. What is the slope of the secant line? Well, they're giving us
two points on this line. It might not be immediately
obvious, but they're giving us the points when x is equal to two, when x is equal to two, what is y? Well y, they tell us, y is
equal to the natural log of x, so in this case it is going
to be the natural log of two, and when x is equal to two plus h, what is y? Well, y is always going
to be the natural log of whatever x is. So it's going to be the
natural log of two plus h. And so these are two points
that sit on the secant line. This happens to be where
the secant line intersects our curve, but these are
two points on the line, and if you know two points on a line, you will then be able to figure
out the slope of that line. Now we can just remind ourselves that a slope is just change
in y over change in x, and so what is this going to be? Well if we view the second
one as our endpoint, our change in y going from ln
of two to ln of two plus h, so our change in y is
going to be our endpoint. So, natural log of two plus
h minus our starting point, or our end y-value minus
our starting y-value. Natural log of two and then our change in x, our change in x is going
to be our ending x-value, two plus h, minus our starting
x-value, minus two, and of course these twos cancel
out, and if we look here it looks like we have a
choice that directly matches what we just wrote. This right over here, natural log of two plus h minus natural log of two over h. Now, if you wanna visualize
this a little bit more, we could draw a little bit,
I'm gonna clear this out so I have space to draw the graph, just so you can really visualize
that this is a secant line. So let me draw my y-axis, and let me draw my x-axis, and y equals the natural
log of x is going to look, so let me underline that, that is going to look something like this. I'm obviously hand drawing it, so not a great drawing, right over here. And so when we have the point two comma natural log of two, which would be, lets say it's over, so if this is two, then this right over here is
the natural log of two, so that's the points two comma
natural log of two, and then we have some other, we just
noted the abstract two plus h, so it's two plus something. So let's say that is two plus h. And so this is going to be the point where we sit on the graph. That's going to be two plus
h comma the natural log of two plus h, and the exercise
we just did is finding the slope of the line
that connects these two. So the line will look
something like that, and and the way that we did
this is we figured out, okay what is our change in y? So our change in, so let's
see, we are going from y equals natural log of two to y equals natural log of two plus h. So our change in y, our change in y is our natural log of two plus h minus natural log of two. Minus natural log of
two, and our change in x? Well we're going from two to two plus h, we're going from two to two
plus h, so our change in x, we just increased by h. We're going from two to two plus h, so our change in x is equal to h. So the slope of the secant line, the slope of the secant line, a secant that intersects
our graph in two points is going to be change
in y over change in x, which is once again exactly
what we have over there.