How do you find the maxima and minima? y=x^2-6x+4?

Question

I really don't know :O

here's an example:

http://www.kumon.ne.jp/english/about/print/k.html

Answer 1

f(x) = x² - 6x + 4
f `(x) = 2x - 6 = 0 for turning point
x = 3 for turning point
f " (x) = 2 , a +ve number, so turning point is a MINIMUM turning point.
f(3) = 9 - 18 + 4 = - 5
Minimum turning point (3, - 5)
There is one turning point only.

Answer 2

The usual means is to take the derivative, in this case y' = 2x -6. Since at either a maximum or a minumum, the slope (i.e., the derivative) must be 0, there is exactly one minimax point, at x = 3. In this case, one can tell by inspection (coefficient of x^2 is positive) that the point is a minimum; more generally, one can take the second derivative y" = 2, note that it is positive, and that means you have a minimum.

Answer 3

First let us take the first derivative,
dy/dx = 2x - 6

set dy/dx = 0, to find the critical points,

0 = 2x - 6
x = 3

Now, to find whether the critical pt is a maxima/minima, find y'' and substitute the pt. If y'' is +ve, it is a minima and if y'' is -ve, it is a maxima.

y'' = d^2(y)/dx^2 = 2
y'' is +ve

So, x = 3, is a Minima.

Answer 4

if you've had calculus then y'=2x-6 = 0 so x=3 is a minimum.
If you haven't had calc then y=(x²-6x+9)+4-9
y=(x-3)² - 5 so a minimum occurs at (3,-5)
We know it's a minimum because b²>4ac so the curve crosses the x axis, but our point of max-min point is below the axis

Answer 5

Differentiate the expression, you get:
dy/dx = 2x - 6

Equate it to 0:
2x - 6 = 0
x = 3

Differentiate it once again:
d/dx (dy/dx) = 2

Since, the result is positive, there is a minima at x = 3 and no maxima.

Answer 6

Use Fermat's theorem

http://en.wikipedia.org/wiki/Fermat%27s_theorem_%28stationary_points%29

Answer 7

you do -b/2a and then plug in that number for x and thats ur max/min...i think