# Minimum value as x varies

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.

## Homework Statement

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

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

## 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

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

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Trail_Builder
what's a derivative?

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.

Homework Helper
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?

turdferguson
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?

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

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

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 :-/

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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)

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

Homework Helper
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.

Trail_Builder
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?

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

where would i go from there?

Homework Helper
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?

Trail_Builder
o yeh oops, lol, my bad

would the minimum value be -m^2 then?