How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for $2 x^{2} + 6 x - 1$ $2x^2+6x-1$ ?

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How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for $2 x^{2} + 6 x - 1$ $2x^2+6x-1$ ?

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Answer 1 · 2015-06-02T12:31:50

$y = 2 x^{2} + 6 x - 1$ $y = 2x^2+6x-1$ is already in standard form

$XXXXX$ $color(white)("XXXXX")$ General standard form for a parabola is $y = a x^{2} + b x + c$ $y=ax^2+bx+c$

Vertex form for a parabola is $y = m (x - a) + b$ $y =m(x-a)+b$
$XXXXX$ $color(white)("XXXXX")$ Note: the $a and b$ $a and b$ are not the same as in the standard form but they are constants.

$y = 2 x^{2} + 6 x + 1$ $y=2x^2+6x+1$
$XXXXX$ $color(white)("XXXXX")$ extract $m$ $m$
$y = 2 (x^{2} + 3 x) + 1$ $y= 2(x^2+3x) + 1$
$XXXXX$ $color(white)("XXXXX")$ complete the square
$y = 2 (x^{2} + 3 x + {(\frac{3}{2})}^{2}) + 1 - \frac{9}{2}$ $y=2(x^2+3x+(3/2)^2) +1 - 9/2$

$y = 2 {(x + \frac{3}{2})}^{2} - \frac{7}{2}$ $y = 2(x+3/2)^2 -7/2$

$y = 2 {(x - (- 3))}^{2} + (- \frac{7}{2})$ $y = 2(x-(-3))^2+(-7/2)$
$XXXXX$ $color(white)("XXXXX")$ which is in vertex form with a vertex at $(- 3, - \frac{7}{2})$ $(-3,-7/2)$

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How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for $2 x^{2} + 6 x - 1$ $2x^2+6x-1$ ?

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How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for $2 x^{2} + 6 x - 1$ $2x^2+6x-1$ ?