Minimum value as x varies

1. Mar 8, 2007

Trail_Builder

hi, I am just tackling some of the harder questions we have to do and I came across this which I have never been told how to solve and my maths teacher too busy teaching the eaiser stuff to other people to expalin this lol, annoying i know, a few of us are in this situation, anyways, hear it is, thnx.

1. The problem statement, all variables and given/known data

The expression x^2 - 2mx has a minimum value as x varies

qu) Find the minimum value of x^2 - 2mx

2. Relevant equations

3. The attempt at a solution

I have no idea, how to approach this. I have tried reasoning logically how it might work, but came nowhere as I dunno how the expression can have a minimum value :S is the power of 2 a clue? A few of us (in my class) have tried but no-one has the fainted idea. hope you can help

thnx

2. Mar 8, 2007

If a function have a maximum or minimum at a certain point, what can you say about the derivative at that point?

Note that this is the PRECALCULUS fourm. Knowledge of derivatives should not be expected
Integral

Last edited by a moderator: Mar 8, 2007
3. Mar 8, 2007

Trail_Builder

what's a derivative?

4. Mar 8, 2007

A derivative is the rate of change of a quantity. A derivative is an instantaneous rate of change: it is calculated at a specific instant rather than as an average over time.

5. Mar 8, 2007

Dick

There is a more elementary way to do problems like this. Complete the square in the expression. Can you handle this? Ie write it as (x-m)^2+SOMETHING. Can you figure out what the SOMETHING is?

6. Mar 8, 2007

turdferguson

You can do this or find the axis of symmetry and vertex. For parabolas, you dont need to use derivatives

7. Mar 8, 2007

Feldoh

By completing the square you're essentially doing that, it will be in vertex form. $$y=a(x-h)^2+k$$

Whatever makes (x-h) equal 0 is your x-coordinate, k is your y-coordinate for a min/max of a quadratic

Or at least I believe what I said to be correct, maybe I'm wrong, dunno :-/

Last edited: Mar 8, 2007
8. Mar 8, 2007

drpizza

if you change the value of "m", with it always be a "happy parabola"?

How do you find the axis of symmetry for the following parabolas?
y=x^2 + 4x -8
y=x^2 + 4x

Then, how would you find the axis of symmetry for your parabola?

Notice that the lowest point for a "happy" parabola is always on this axis. (likewise for a "sad" parabola)

9. Mar 9, 2007

dragonlorder

the lowest point is min value let say ax^2+bx+c
the lowest point x-coordinate is /frac{b}{-2a}

10. Mar 9, 2007

Gib Z

Nice try at the latex dragonlorder, welcome to Pf. You forgot [ tex ] brackets though. As said before, the x value at the lowest point is $\frac{-b}{2a}$, to get the lowest value, they are asking for the Y value! So dont forget to substitute your x value back into your equation and get the y value.

11. Mar 9, 2007

Trail_Builder

would it be (x-m)^2 + m^2?

where would i go from there?

12. Mar 9, 2007

Dick

I would say x^2-2mx=(x-m)^2-m^2. Now (x-m)^2 is always greater than or equal to zero, yes? So what is the minimum value?

13. Mar 9, 2007

Trail_Builder

o yeh oops, lol, my bad

would the minimum value be -m^2 then?

14. Mar 9, 2007

Dick

You are so right.

15. Mar 9, 2007

Trail_Builder

thnx for all the help buddy :D