Finding a slope at a point on quadratic (intuition of limit)

In summary, the author attempted to find the slope and equation of a line that passes through -2,8. He understood the process up to the point where he set h=0, but is stuck on how to find the slope of the line when h is not 0. He finds clarity by reading a book on the subject that uses the concept of hyperreal numbers.
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Homework Statement


Find the slope of ##y=x^2+4## at (-2,8) and the equation for this line.

Homework Equations

The Attempt at a Solution



This problem is intended to give an intuition on how limits work and I think I get the general idea.
If we want to find the rate of change (or slope) of some point on a function f(x), we essentially "close in" on that point, a, with another point until they're virtually the same point. This creates a tangent line to the point on the graph of the function and we can take the slope of that line and thus know the rate of change of f(x) at that point, a. There are other specifics, but I'm just looking for an answer as to what is happening on this worksheet (attached).

I understand everything up to the point until we set h=0. If we want to close in on the point (-2,8), wouldn't we want to approach that x-value? And since we haven't seen how to take a limit yet, set h equal to -2 rather than 0?
 

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  • #2
Look at the solution. h is the distance away from point x = -2, not x. So when h is set equal to zero you close in point x = -2 as you say.
 
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  • #3
Actually I just realized what it is. By setting h=0, we are "approaching" -2 as I said. Setting h=-2 would make it -4. Not sure what I was thinking there.

Now a new question then. By setting h=0, we have the exact point we are trying to take the slope of, (-2,8). I feel like I'm looking at this the wrong way. We did those calculations to arrive at the same point we were told to find the slope at.

Edit: You beat me to it! Thank you.
 
  • #4
Are you stuck? I can't tell.
 
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  • #5
I was, but I got it now. It was more of a conceptual thing. The book I'm reading out of doesn't explain things very well so I looked at another one and it used the idea of "hyperreal" number to explain it. Makes much more sense. Thank you.
 

1. What is the formula for finding the slope at a point on a quadratic function?

The formula for finding the slope at a point on a quadratic function is given by the derivative of the function evaluated at that point. In other words, it is the instantaneous rate of change of the function at that specific point.

2. How is the slope at a point on a quadratic function related to the concept of a limit?

The slope at a point on a quadratic function is related to the concept of a limit because the derivative, which represents the slope, is defined as the limit of the average rate of change as the change in x approaches zero. In other words, the slope at a point is the limit of the function's rate of change at that point.

3. Why is finding the slope at a point on a quadratic function important?

Finding the slope at a point on a quadratic function is important because it allows us to determine the rate of change of the function at that point. This information is useful in many real-world applications, such as determining the velocity of an object or the growth rate of a population.

4. Can the slope at a point on a quadratic function ever be undefined?

Yes, the slope at a point on a quadratic function can be undefined if the function has a vertical tangent at that point. This occurs when the derivative of the function is equal to zero, meaning the function has a critical point or point of inflection at that point.

5. How can we use the slope at a point on a quadratic function to determine its concavity?

The slope at a point on a quadratic function can be used to determine its concavity, or the direction in which the function is curving. If the slope is positive, the function is concave up, and if the slope is negative, the function is concave down. This information can be helpful in understanding the overall shape and behavior of the function.

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