What is the vertex of $y = 3 x^{2} + 6 x + 1$ $y=3x^2 + 6x + 1$ ?

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Answer 1 · 2015-11-17T17:10:35

$(- 1, - 2)$ $(-1, -2)$

Derivate the function and calculate $y'(0)$ to find where the the slope is equal to $0$ .

$y = 3x^2 + 6x + 1$
$y' = 2 * 3x^(2-1) + 1 * 6x^(1-0)$
$y' = 6x + 6$

Calculate $y'(0)$ :
$y'(0) = 0$
$6x + 6 = 0$
$6x = -6$
$x = -1$

Put this $x$ value into the original function to find the y-value.
NOTE: Put it in $y$ , not $y'$ .

$y = 3*(-1)^2 + 6*(-1) + 1$
$y = 3 * 1 - 6 + 1$
$y = 3 - 6 + 1 = -2$

The vertex is at $(-1, -2)$

What is the vertex of y=3x^2 + 6x + 1 y=3x2+6x+1 y=3x^2 + 6x + 1 ?