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

  • Thread starter opus
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opus
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Homework Statement


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

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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
kuruman
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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.
 
  • #3
opus
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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
verty
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Are you stuck? I can't tell.
 
  • #5
opus
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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.
 

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