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September 26th, 2014, 08:53 AM   #1
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Equation jump in graphical algebra equation

So, recently I was doing a math involving graphs but I had some difficulty solving it. I asked my Mathematics teacher about this and he wrote down the equation of the graph and then he did something completely unexpected:
This step I did not get. When I asked him to explain he said something like the power of x was multiplied by it's coefficient and the same for 3x and -3.
Next in the equation he did this:
dy/dx=2x+3 (he said dy/dx was the gradient and I get this part. From the question, we know the gradient is 1.)

Would be glad if anyone explains this. I'm adding the question:

"The variables x and y are connected by the equation y=x^2+3x-3.
Some corresponding values of x and y are given in the following table
x|-5|-4|-3|-2|-1 |0 |1|2|

y|7 | 1| -3|-5|-5|-3|1|7|

a) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 2 units on the y-axis, draw the graph of y=x^2+3x-3 for -5<x<2

b) By drawing a suitable tangent to your curve, find the coordinates of the point at which the gradient of the tangent is equal to 1.

#b) is the one I'm having problem with.

Note: It's not like I didn't pay attention when he explained, so please don't mention that. More than half the class didn't get what he said and he explained more than like 10 times so it would be awkward if we said that we still don't get it :/

Last edited by skipjack; September 27th, 2014 at 06:26 AM.
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September 26th, 2014, 02:07 PM   #2
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Originally Posted by DarkX132 View Post
he wrote down the equation of the graph and then he did something completely unexpected:
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 $(x,y(x))$.
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September 26th, 2014, 03:23 PM   #3
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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.
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September 26th, 2014, 10:15 PM   #4
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Oh I get it now Thanks
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