Secant line with arbitrary point (with simplification) | AP Calculus AB | Khan Academy

A secant line intersects the graph of f of x, which is equal to x² + 5x, at two points with x-coordinates 3 and T, where T does not equal 3. 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. But anyway, we want to find the slope of the secant line. They essentially give us two points on the secant line. They tell us what x is at each of those two points, and then if we know what x is, we're able to figure out what f of x is at each of those points.

So we can make a little table here. We know x, and we know f of x. When x is equal to 3, what is f of x? Well, it's going to be 3² + 5 * 3. This is going to be 9 + 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 T² + 5T. So we have two points now that are on this line. This is a secant line; it intersects our function twice, so it has these two points on it.

Now, 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 endpoint, the second one with the T's as our endpoint, it's going to be that minus that.

So it's going to be T² + 5T minus 24. In our denominator, our ending x minus our starting x is going to be T - 3. 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 - 3? Alright, so in the numerator, let’s see. -3 * 8 is -24, plus positive 8 is 5. So we can rewrite this as (T + 8)(T - 3).

And then we could say this is going to be equal to, if we cancel out the T - 3's or we divide the numerator and the denominator by T - 3, it's going to be equal to T + 8.

Now, if we wanted to be really strict mathematically, 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 T's except where T equals 3. This thing right over here is defined at T equals 3. In fact, when T equals 3, this expression is equal to 11. But this thing up here was not defined at T equal 3.

So if you wanted to be particular about it, if you want this expression to be the exact same thing, you would say, you would say for T does not equal 3. Now, this can take the same inputs as this one right over there, but I guess they're assuming where T does not equal 3.

So this, you could view this as maybe a little bit redundant, but this would be the slope of the secant line in terms of T.