Main content

## Differential Calculus (2017 edition)

### Unit 5: Lesson 4

Secant lines- Slope of a line secant to a curve
- Secant line with arbitrary difference
- Secant line with arbitrary point
- Secant lines & average rate of change with arbitrary points
- Secant line with arbitrary difference (with simplification)
- Secant line with arbitrary point (with simplification)
- Secant lines & average rate of change with arbitrary points (with simplification)
- Secant lines: challenging problem 1
- Secant lines: challenging problem 2

© 2022 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Secant line with arbitrary point (with simplification)

Sal finds and simplifies the expression for the slope of the secant line between x=3 and x=t on the graph of y=2x²+5x.

## Video transcript

- [Voiceover] A secant
line intersects the graph of f of x is equal to
x squared plus five x at two points with x
coordinates three and t where t does not equal three. What is the slope of the
secant line in terms of t? Your answer must be fully
expanded and simplified. And my apologies ahead
of time if I'm a little out of breath. I just tried to do some
exercises in my office to get some blood moving
and I think I'm still a little out of breath. Anyway, so we want to
find the slope of the secant line and they essentially give us two points on the secant line. They tell us what x is at
each of those two points and if we know what an x is, we are able to figure out what f of x is at each of those points. So we could make a little table here. We know x and we know f of x. So when x is equal to three what is f of x? Well it's going to be three squared plus five times three. Well this is going to be nine plus 15 which is 24. So this is going to be 24 and when x is equal to t what is f of t? Well it is going to be
t squared plus five t. And so we have two points
now that are on this line, this is the secant line. It intersects our function twice so it has these two points on it. So we just have to find
our change in y between these two points. Change in y and our
change in x, change in x. And I'm assuming that
y is equal to f of x. So our slope of our secant line is change in y over change in x. Our change in y, if we
view this as our end point, the second one with the
t in it is our end point is going to be that minus that. So it's going to be t squared plus five t minus 24 and then in our denominator our ending x minus our starting x is
going to be t minus three. Now they tell us our answer must be fully expanded and simplified. So maybe there's a way to
simplify this a little bit. Let's see, can I factor
the top into something that involves a t minus three? All right, so in the numerator, let's see, negative three times positive eight is negative 24. Negative three plus
positive eight is five. So we can rewrite this as t plus eight times t times t minus three. And so we could say this
is going to be equal to if we cancel out the t
minus three or we divide the numerator into the
denominator by t minus three it's going to be equal to t plus eight. Now if we wanted to be really strict, mathematically strict, this expression isn't
exactly the same as our original expression right over here. What makes them different? Well they're going to
be true for all of t's except where t equals three. This thing right over here
is defined at t equals three. In fact when t equals three
this expression is equal to 11. But this thing up here was
not defined at t equals three so if you wanted to be particular about it you want this expression
to be the exact same thing you would say four t does not equal three. Now this can take the
same inputs as this one right over there. But I'm assuming where
t does not equal three. So you could view this
as maybe a little bit redundant. But this would be, this is the slope of the
secant line in terms of t.