How do I turn a parabolic curve into a straight line?

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SUMMARY

The discussion focuses on transforming a parabolic curve, specifically the equation y = x², into a linear representation at a given point. The key method involves differentiation to find the gradient, which is calculated as 2x. At the point x = 4, the gradient is determined to be 8, leading to the linear equation y = 8x + 12 using the point-slope formula y - y1 = g(x - x1). Understanding differentiation is essential for accurately performing these transformations.

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  • Differentiation and its application in calculus
  • Understanding of parabolic equations, specifically y = x²
  • Knowledge of the point-slope form of a linear equation
  • Basic graphing skills for visual representation
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Students and educators in mathematics, particularly those studying calculus, as well as anyone interested in understanding the relationship between parabolic curves and linear equations.

twenty5
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sooper easy question :)

uhmm, say I have half of a parabolic curve, how do I turn it into a straight line?

what I'm thinking...

y = x2
root of y = x? but I leave the x as it is?

hmm hard to explain... let me try explaining a bit

(x,y)
(4,16)

okay? ok now to make it straight, I would do root of 4 and leave x as 16 and then just plot this on graph?

result

(x,y)
(2,16)

or would I do the root of 16 as well
 
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I have to ask, what are you trying to do?

"uhmm, say I have half of a parabolic curve, how do I turn it into a straight line?"

That doesn't really make sense.. Are you trying to find the gradient of the curve at a point?
 


Yeah i think he is;

First off you must differentiate, I'm not sure if you know what this means so i'll put a simple version down anyway;

y = x²

So CHANGE in y / CHANGE in x (gradient) = 2x (bring the power down)

Then insert the x figure

so say at the point x = 4, the gradient would be 8. (If you don't understand differentiation, look it up, it is KEY!! to understanding / doing any of these questions) http://en.wikipedia.org/wiki/Derivative

Then you must use the formula

y - y1 = g ( x - x1) Where you insert G gradient, y1 & x1, the points so

y - 16 = 8 (x - 4)

y = 8x + 12 Would be the equation for the line at the point x = 4

I can understand why you might find all that a bit hard to understand, if you still can't pickup on some stuff here post again! You must understand these problems! Peace! :P
 

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