how do i find the minimum

Question

how do I find the minimum of y = x^2 - 4x - 32?

 

I think it starts with y = a(x)^2 + bx + c a > 0

 

 

then,  y = - b/2a

 

 

Answer 1

The function describes a parabola that opens upward.  There are a variety of methods to find the minimum.

 

1) Find the zeros of the equation.  The minimum will occur halfway between the two x-coords zeros since the function is symmetric about a vertical axis.

 

y=x^2-4x-32 = (x-8)(x+4)

The two X-coord zeros (where y=0) are x=8 & x=-4

Therefore, the min occurs at X=2, Y=-36 (plugging X=2 into the equation)

 

2) X=-b/2a does indeed provide the minimum for a parabola and is an easy way to calculate it when presented with standard form Y=aX^2+bX+c.  Xmin=-b/2a = (4/2) =2 (same result)

A word of caution: If a<0, the answer will be a MAXIMUM and not a minimum since the parabola will open downward.  That's why calculus is the preferred method for arbitrary functions.

 

3) Calculus provides the best way to find the extremal points (in general).  By taking the derivative (slope) and setting equal to zero (slope is zero at both minimum and maximum.  The sign of 2nd deriv identifies which it is)

 

y=x^2-4x-32

y'=2x-4   ------>  2x=4  (setting the deriv equal to zero) -----> X=2   (same result)

Answer 2

Hi again L;

y = x² - 4x - 32

As you already know, this is a parabola.  The x² is positive.  Therefore, it is opening upward.  The minimum corresponds to the line-of-symmetry.

This is NOT...

y=-b/2a

This is...

x=-b/2a

x=[(-(-4)]/[(2)(1)]

x=4/2

x=2

y = x² - 4x - 32

y=(2)²-(4)(2)-32

y=4-8-32

y=4-40

y=-36

The lowest point is (2,-36).

 

An alternate approach is to consider the fact that the minimum corresponds to the point at which the change of y is zero.  Therefore, we can take the first derivative of the equation and calculate this...

y = x² - 4x - 32

0=2x-4

4=2x

2=x