If #x^2+bx+40# has a turning point at #x=-7# then what is #b# ?

1 Answer
May 30, 2015

The turning point will occur where the derivative is zero:

#d/(dx)(x^2+bx+40) = 2x+b#

Since we want this to be zero when #x=-7#, we solve

#0 = 2(-7)+b = -14+b#

Add #14# to both ends to get:

#b = 14#

Alternatively, notice that

#(x+b/2)^2 = x^2+bx+b^2/4#

So

#x^2+bx+40 = (x+b/2)^2+(40-b^2/4)#

This will take its minimum value when:

#(x+b/2)^2 = 0#, that is when #x = -b/2#.

If this is where #x=-7#, then we get #-7 = -b/2#, hence #b=14#.