how do I identify the minimum

Question

using y = x^2 - 4x -32 how do I identify the minimum?    how do I find the x-intercepts by factoring?

 

please show steps.

 

 

appreciate your help!

Answer 1

 Minimum value of a quadratic   aX^2 + bx +c  a>0

 

  Maximum for the  a<0 is always at point :

 

( X, Y) =  ( -b/ 2a, f( -b/ 2a) = b^2 - 4ac )

                                              4 a^2

 

     Y = X^2 - 4x - 32

 

      X min = 4/ 2 = 2

 

     Y min = ( 2^2) - 4 ( 2) - 32= -36

 

 

 

 

Answer 2

Hi L;

y = x² - 4x -32

Because x² is positive, we know this is a parabola opening upward.  The line of symmetry is the minimum.

The equation is in the standard format of...

y=ax²+bx+c

The line of symmetry is...

x=-b/2a

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

x=4/2=2

y=[2²]-[(4)(2)]-32

y=4-8-32

y=4-40

y=-36

(2,-36)