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## Calculus 1

### Course: Calculus 1 > Unit 2

Lesson 2: 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

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# Secant line with arbitrary difference (with simplification)

Sal finds and simplifies the expression for the slope of the secant line between x=4 and x=4+h on the graph of y=2x²+1.

## Want to join the conversation?

- At the end, is in necessary to specify that h cannot equal 0? Or can h = 0 when we find the slope of the secant line to be 2h + 16?(13 votes)
- Is it necessary? It depends. If you are in a classroom, your teacher may tell you h cannot equal 0 once (because if h=0 then it would be on the same point as x=4, then there will be no secant line or there is the problem with dividing by 0 if h=0) and never repeat it again with the assumption that their students understood that h cannot equal 0.

But in situation like this, an online video posting for a broader range of students who can access it at all different time, then yes, it is necessary. Otherwise, they may ask, what if h=0 and we can't divide by 0?

But it is always a good habit/practice to include it regardless of situations. So is it necessary? It depends on situations. Should it be specified? Absolutely.(19 votes)

- Is the slope of a secant line same as the average of the slopes of the tangent lines to the tow points of that secant line?(6 votes)
- Nice question! This statement is not necessarily true.

Consider the function f(x) = x^3 and the points (0, 0) and (1, 1) on the graph of this function.

The slope of the secant line is [f(1)-f(0)] / (1-0) = (1-0)/(1-0) = 1.

Since f'(x) = 3x^2, the slopes of the tangent lines at these points are f'(0) = 0 and f'(1) = 3. So the average of the slopes is (0+3)/2 = 1.5, which does not equal 1.(8 votes)

- Could someone please explain what Sal describes at 1.17 referring to 16+8h+h^2? How did Sal get to 8h?(7 votes)
- Using this method can we also assume that the slope of the tangent line at 4 is 16? Because as we bring h to 0 the two points become the same (making a tangent line) and 2(0) + 16= 16. Or am I mistaken?(4 votes)
- This is correct. In fact, when defining the derivative as a limit in the next videos, we basically do what you describe here.(4 votes)

- I understand this 90%. However , for the 2h^2 + 16h + 33 -33, how did he get the "-33"?(2 votes)
- You're taking the difference between the two y coordinates. So it's like saying "find the difference between the second y coordinate and the first y coordinate." We can express that mathematically by saying (the second y coordinate)-(the first y coordinate)=(2h^2+16h+33)-(33). Hope this helps. Good luck!(6 votes)

- which is more simplified; 2h+16 or 2(h+8) ?(3 votes)
- 2(h+8) is more simplified than 2h+16. The expression 2h+16 is composite, meaning it is composed of sub-units 2 and (h+8) multiplied together. The process of simplification renders an expression in it's least possible elements.(1 vote)

- At the end of the video, when we have figured out the slope, can 2h+16 be simplified to h+8? Because 2h and 16 are both divisible by 2, and the problem asked us to fully simplify this.(1 vote)
- there is nothing to simplify. We can only take the two out if it was (2h+16)/2 which it isn't. and 2(h+8) and 2h+16 are both simplified enough.(1 vote)

- What does a slope of 2h+16 even mean? It makes sense for the slope to be a constant, but I can't imagine what a slope with a variable would look like.(1 vote)
- Yes, the slope will be constant for a straight line, but
`y=2x²+1`

is a curve – a parabola to be exact (if that doesn't mean anything to you I recommend going through the material on "conics" https://www.khanacademy.org/math/algebra2/intro-to-conics-alg2).(1 vote)

- But wait, as h approaches zero won't the slope approach 16? Is that logical?(1 vote)
- As h approaches 0, we get x =4. So, applying limit into the slope implies that the slope is a constant (the derivative) for a particular x-coordinate. We can understand that what is happening is just the slope is constant for a single x value.(1 vote)

- what is between the difference secant line and slopes(1 vote)
- A secant line is an object. We say a line is
*secant*to a curve when it touches the curve in two places. For example, a chord in a circle is secant to the circle.

The slope of a line is measure of how "tilted" the line is. A horizontal line has a slope of 0. As we rotate the line counterclockwise, the slope increases until it becomes infinite (and so, undefined) when the line is vertical. If we keep rotating it, the slope become negative.(1 vote)

## Video transcript

- [Voiceover] A secant line intersects the curve y = 2x squared + 1 at two points with x-coordinates 4 and 4 +
h, where h does not equal 0. What is the slope of the
secant line in terms of h? Your answer must be fully
expanded and simplified. 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. We have x and then we
have y = 2x squared + 1. So we know that when x = 4, what is y going to be equal to? It's gonna be 2(4 squared) + 1 which is the same thing as 2(16) +1 which is the same thing as 32 + 1 so it is going to be 33. What about when x = 4 + h? Well it's going to be 2(4 +h)squared + 1. That's going to be 2 times, let's see, (4 + h)squared is going to be 16 + 8h + h squared and then we have our + 1 still and if we distribute the 2 that's going to get us to
32 + 16h + 2h squared + 1 and then we add the 32 to the 1 and actually I'm gonna
switch the order a little bit so I have the highest degree term first so it's going to be 2h squared + 16h and then + 32 + 1 is 33. We have these two points. We have one point (4, 33) and we have the other point
(4+h, 2h squared + 16h + 33) and we just have to find the slope between these two points because the secant line
contains both of these points. How do we find the slope of a line? We do change in y / change in x. What's our change in y? If we view this as the end point and this as the starting point, our change in y is going
to be this minus that. It's going to be 2h
squared + 16h + 33 - 33, those two are going to
cancel each other out, and then over, what's our change in x? If we ended at 4 + h but then we started at 4 so it's gonna be 4 + h - 4. These two cancel each with each other and we are left with 2h squared + 16h / h. We can divide everything in the numerator and denominator by h and what are we going to get? This is going to be 1. That's just a 1. This is just an h. We have 2h + 16 / 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 have to think about well the secant line
contains the point (4,f(4)) or 2 times 4 squared + 1 right over here and, well I didn't call this f(x) but I think you get the idea, and then when x is 4 + h this is going to be y and we just found the slope
between these two points.