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msbb45 · Oct 17, 2009, 06:52 AM

f(x)=-2x^2+2x+1

x-coordinate
y-coordinate

the equation of line of symmetry
the max/min

Unknown008 · Oct 17, 2009, 09:17 AM

Well, the usual method is to complete the square, then sketch your graph. This will automatically tell you whether the turning point is a max or min.

The second method is to find the derivative of your function, then equate that to zero. This gives you the x coordinate of the vertex. Replace that value of x in your function, to get the value of the y coordinate. The nature of the turning point is then given by the second derivative of the function.

Nhatkiem · Oct 17, 2009, 10:44 PM

Sorry to say, but there are much easier methods than to find its for evaluating the position of the vertex. Since it is only a second ordered equation, it is parabolic in nature and finding the vertex won't take much effort at all :D!

If we look at f(x)=-2x^2+2x+1

We see that -2 is the leading coefficient for x^2
2 is the leading coefficient for x
and 1 is the constant

It is in the form Ax^2 +Bx+C!

the x coordinate of the vertex for parabolic functions is
x=-b/2a

In this situation we have

x= -2/2(-2) = 1/2

Plus in this value of 1/2 for x, and we get

f(1/2) = -2(1/2)^2+2(1/2)+1 = -1/2+2 = 3/2

Unfortunately you can't avoid derivatives if you want the most quick route to determine max/min! f"(1/2) > 0 (implies the function is increasing) then we have a minimum, if f"(1/2)<0 (implies the function is decreasing), we have a minimum.

Unknown008 · Oct 18, 2009, 02:39 AM

Actually, the method you described is the same as for completing the square, which is from where you get the formula for the x coordinate of the vertex to be x = -b/2a.

See the general quadratic?

$ax^2 + bx + c = 0$

The completed square from is:

$a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$

You get the x coordinate by equating that in the brackets to zero:

$x + \frac{b}{2a} = 0$

$x = - \frac{b}{2a}$

And you get the y coordinate as the values after the brackets, that is:

$y = - \frac{b^2}{4a} + c$

See?

I prefer that one does the complete procedure, knowing where which and which formula comes from. :)

Nhatkiem · Oct 18, 2009, 11:23 AM

Thank, I do see that. I'll try to keep in mind that I shouldn't solve the problems, only show how it could be done.

Unknown008 · Oct 19, 2009, 09:24 AM

Yes, it's better that way. People most of the time copy down the answers when it is given to them, and then when they get similar problems, they get stuck again. That is worse when they have to solve the problem in an exam, or a test.