Question

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

Answer 1

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`.