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

May 30, 2015

The turning point will occur where the derivative is zero:

$\frac{d}{\mathrm{dx}} \left({x}^{2} + b x + 40\right) = 2 x + b$

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

$0 = 2 \left(- 7\right) + b = - 14 + b$

Add $14$ to both ends to get:

$b = 14$

Alternatively, notice that

${\left(x + \frac{b}{2}\right)}^{2} = {x}^{2} + b x + {b}^{2} / 4$

So

${x}^{2} + b x + 40 = {\left(x + \frac{b}{2}\right)}^{2} + \left(40 - {b}^{2} / 4\right)$

This will take its minimum value when:

${\left(x + \frac{b}{2}\right)}^{2} = 0$, that is when $x = - \frac{b}{2}$.

If this is where $x = - 7$, then we get $- 7 = - \frac{b}{2}$, hence $b = 14$.