How do you find the stationary points of the function $y = x^{2} + 6 x + 1$ $y=x^2+6x+1$ ?

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Answer 1 · 2018-04-10T12:03:02

$(- 3, - 8)$ $(-3,-8)$

The stationary points of a function are when $\frac{d y}{d x} = 0$ $dy/dx=0$

$y = x^{2} + 6 x + 1$ $y=x^2+6x+1$

$\frac{d y}{d x} = 2 x + 6$ $dy/dx=2x+6$

$\frac{d y}{d x} = 0 = 2 x + 6$ $dy/dx=0=2x+6$

$x = - \frac{6}{2} = - 3$ $x=-6/2=-3$

${(- 3)}^{2} + 6 (- 3) + 1 = 9 - 18 + 1 = - 8$ $(-3)^2+6(-3)+1=9-18+1=-8$

Stationary point occurs at $(- 3, - 8)$ $(-3,-8)$

How do you find the stationary points of the function y=x^2+6x+1y=x2+6x+1y=x^2+6x+1?