How do you find the minimum and maximum value of #y=(x+7)(x+3)#?

1 Answer
Feb 21, 2017

minimum #->(x,y)=(-5,-4)#
maximum #-> y=+oo#

Explanation:

If you multiply out the brackets you have the general form of:
#y=ax^2+bx+c# in this case #a=1# giving just #+x^2#

As this is positive the graph is of general shape #uu#
Thus the vertex is a minimum.

The vertex is #1/2# way between the x-intercepts.

Set:#" "y=0=(x+7)(x+3)#

Thus

#x+7 =0 => x=-7#
#x+3=0=>x=-3#

So #x_("vertex")=(-3-7)/2=-5#

Substitute #x=-5# to give:

#y_("vertex")=(-5+7)(-5+3)= 2xx(-2)=-4#

Thus the minimum #->(x,y)=(-5,-4)#

As the graph is of form #uu# then the maximum is #y=+oo#