# Thread: Four step method (two point method)

1. ## Four step method (two point method)

Hey guys
I understand how to get the slope (getting the imaginary number, etc) but I've only ever done it when x = 1 or -1. How do I do it when x = -3?
I have the question: y = x^2, while x = -3.

I tried doing it, but I got a slope of 2...

Thanks for any help.

2. ## Re: Four step method (two point method)

Originally Posted by eden19
Hey guys
I understand how to get the slope (getting the imaginary number, etc) but I've only ever done it when x = 1 or -1. How do I do it when x = -3?
I have the question: y = x^2, while x = -3.

I tried doing it, but I got a slope of 2...

Thanks for any help.
We might benefit from a bit more of a clue as to what your question is rather than the rather cryptic text above.

CB

3. ## Re: Four step method (two point method)

Using the two point formula, find the derivative of the function and the instantaneous rate at the x-coordinate provided.

1. y = x^2 at x = -3

4. ## Re: Four step method (two point method)

Originally Posted by eden19
Using the two point formula, find the derivative of the function and the instantaneous rate at the x-coordinate provided.

1. y = x^2 at x = -3
Well, I, for one, was puzzled by your reference to "getting the imaginary number, etc." There are no imaginary numbers involved in this problem. I wonder if you were referring to finding the limit?

The "two point formula" for slope is (y2- y1)/(x2- x1) where (x1, y1) and (x2, y2) are two points on the curve. Here, y= x^2 so y1= x1^2, y2= x2^2. (y2- y1)/(x2- x1)= (x2^2- x1^2)/(x2- x1). Can you simplify that? (Factor the numerator.)

To find the "derivative of the function and the instaneous rate at they coordinate provided" (they are the same thing), take the limit as x1 goes to x2. Finally set the value of x2 to 3.
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