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

A secant line intersects the curve ( y ) is equal to ( 2x^2 + 1 ) at two points with ( x ) coordinates ( 4 ) and ( 4 + h ), where ( h ) does not equal zero. What is the slope of the secant line in terms of ( h )? Your answer must be fully expanded and simplified.

So we know the two points that are on the secant line. It might not be obvious from how they wrote it, but let's make a little table here to make that a little bit clearer.

So we have ( x ) and then we have ( y ), which is equal to ( 2x^2 + 1 ). And we know that when ( x ) is equal to ( 4 ), well what is ( y ) going to be equal to? Well, it's going to be ( 2 \times 4^2 + 1 ), which is the same thing as ( 2 \times 16 + 1 ), which is the same thing as ( 32 + 1 ). So it is going to be ( 33 ).

What about when ( x ) is equal to ( 4 + h )? When ( x = 4 + h ), well it's going to be ( 2 \times (4 + h)^2 + 1 ). Well, that's going to be ( 2 \times (4 + h)^2 ) which is going to be ( 16 + 2 \times 4h ), so it's going to be ( 8h + h^2 ), and then we have our plus ( 1 ) still.

If we distribute the ( 2 ), that's going to get us to ( 32 + 16h + 2h^2 + 1 ). Then we add the ( 32 ) to the ( 1 ) and actually I'm going to switch the order a little bit so I have the highest degree term first. So it's going to be ( 2h^2 + 16h ) and then plus ( 32 + 1 ) is ( 33 ).

So we have these two points: we have ( (4, 33) ) and we have the other point ( (4 + h, 2h^2 + 16h + 33) ). We have to find the slope between these two points because the secant line contains both of these points.

So how do we find the slope of a line? Well, we do a change in ( y ) over change in ( x ). So it's our change in ( y ). Well, if we view this as the endpoint and this is the starting point, our change in ( y ) is going to be this minus that. So it's going to be ( 2h^2 + 16h + 33 - 33 ).

Those two are going to cancel each other out. Now, over what is our change in ( x )? Well, if we end it at ( 4 + h ), but then we started at ( 4 ), it's going to be ( (4 + h) - 4 ).

These two cancel with each other and we are left with ( 2h^2 + 16h ) over ( h ). Well, we can divide everything in the numerator and denominator by ( h ). What are we going to get? This is going to be ( 2h + 16 ) over ( 1 ) or just ( 2h + 16 ).

And we're done! This is the slope of the secant line in terms of ( h ). Once again, we just had to think about the secant line containing the points ( (4, f(4)) ) or ( 2 \times 4^2 + 1 ) right over here, and then ( 4 + h ).

Well, I didn't call this ( f(x) ) but I think you get the idea, and then when ( x ) is ( 4 + h ), well this is going to be ( y ), and we just found the slope between these two points.