Lesson Shortcut in finding the vertex of any parabola

Consider any parabola.
A tangent of a curve is a line that passes through one point only(tangent means 'to touch')
And, the vertex of any parabola is the point where the only HORIZONTAL LINE PASS THROUGH!
The thing is, it requires DIFFERENTIATION!
So, I'll teach you a basic differentiation rule commonly used in parabolas----The EXPONENT RULE!
If y=x^n

, then,
_______________________________________ dy/dx=nx^(n-1)

This statement is read as follows:
If y is defined as x raised to the power of n, then, the DIFFERENTIAL of y (written as y'), is x, raised to the power of n-1 and multiplied to n.
___________ y=x^n

Look at the equation above.
The easy way to remember the differential is to write first dy/dx!
___________ dy/dx

Then from the equation y=x^n

, since n is the exponent, let's bring it down!
___________ dy/dx=n

Then, write x!
___________ dy/dx=nx

Since n is already brought down, let's raise x to one less than n or n-1

.
___________ dy/dx=nx^(n-1)

There, you're done!
This is kinda hard to explain. I'll redefine it ONE MORE TIME!

Example: y=x^3

Find dy/dx
It is obvious that n=3. So,
dy/dx=3x^(3-1)

Yahoo!
Now, no kiddin', we're going to apply it now.
If a parabola is given to be, y=2x^2-4x+9

Before I can forget,dy/dx, in the equation represents the SLOPE OF THE CURVE in a given point
Finding y',
dy/dx=(2(2x^(2-1)))-(1(4x^(1-1)))+0

The DIFFERENTIAL OF A CONSTANT IS ZERO!!!
dy/dx=4x-4

Since the horizontal line has a SLOPE of zero,
0=4x-4

This means that the X-COORDINATE of the VERTEX is 1.
If x=1,then, to find y, we refer to the original equation--- y=2x^2-4x+9

Thus,

Therefore, the VERTEX calculated from the SHORTCUT is (1, 7)
Wanna see a proof? Look below!
graph(900,900,-10,10,-10,10,2x^2-4x+9,7)

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