My Math Forum Equation jump in graphical algebra equation

 Algebra Pre-Algebra and Basic Algebra Math Forum

September 26th, 2014, 02:07 PM   #2
Senior Member

Joined: Dec 2013
From: Russia

Posts: 327
Thanks: 108

Quote:
 Originally Posted by DarkX132 he wrote down the equation of the graph and then he did something completely unexpected: y=x^2+3x-3 or,y=2x+3
He probably meant $\dfrac{dy}{dx}=2x+3$, as you write later. The expression $\dfrac{dy}{dx}$ is the derivative, and you can read more about it in Wikipedia (see especially the "Derivatives of elementary functions" section). It is also the slope of the tangent line at point $(x,y(x))$.

 September 26th, 2014, 03:23 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2296 Math Focus: Mainly analysis and algebra Could your teacher have written $$y^\prime = 2x + 3$$ This is another way of writing $$\frac{\mathbb d y}{\mathbb d x} = 2x + 3$$ As to understanding it, I assume you haven't met the theory of limits yet, so the best bet will be the way Isaac Newton developed it. We can approximate the gradient of a curve $y=f(x)$ by taking any two points $(x,f(x))$ and $(x+c,f(x+c))$ on the curve. We can then join the two points with a straight line. The gradient of the curve (at $x$) is approximately the gradient of the line. And the gradient of the line is$$\frac{\text{change in y}}{\text{change in x}} = \frac{f(x+c)-f(x)}{(x+c) - x} = \frac{f(x+c)-f(x)}{c}$$ Now, for large $c$, this approximation is poor, but the smaller we make $c$, the better the approximation becomes. So let's make $c$ really, really, small. As small as you can imagine. Clearly, the approximation is now really, really good. So let's see what our approximation gives $$\frac{f(x+c)-f(x)}{c} = \frac{\left( (x+c)^2 + 3(x+c) - 3\right) - \left( x^2 +3x -3\right)}{c} = \frac{\cancel{x^2} + 2xc + c^2 + \cancel{3x} + 3c - \cancel3 - \cancel{x^2} - \cancel{3x} + \cancel3}{c} = 2x + 3 + c$$ And now comes Newton's fudge. He says that because $c$ is really, really small, we can ignore it. So the gradient of the curve at any point $x$ is$$2x+3$$ In the years since Newton developed this method, it has been refined and made rigorous, but we still use his 'difference quotient' as the starting point. Thanks from greg1313
 September 26th, 2014, 10:15 PM #4 Newbie   Joined: Sep 2014 From: Bangladesh Posts: 4 Thanks: 0 Oh I get it now Thanks

 Tags algebra, equation, graphical, jump

,

,

,

# in a graph how do you draw 2 cm to 2units y axis

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post fantom.1040 Algebra 5 June 11th, 2013 12:41 AM emohbe Algebra 2 December 13th, 2010 08:30 PM iloatheparabolas Algebra 1 August 13th, 2008 05:14 PM albertjohansson337 Algebra 1 June 28th, 2008 06:44 PM johnny Algebra 1 December 26th, 2007 09:30 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top