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How do you find the VERTEX of a parabola $y = x^{2} + 7 x + 12$ $y=x^2+7x+12$ ?

1 Answer

Sep 8, 2015

Use the method of completing the square to convert to vertex form and get the vertex at $(- \frac{7}{2}, - \frac{1}{4})$ $(-7/2, -1/4)$

Explanation:

Vertex form: $y = m {(x - a)}^{2} + b$ $y=m(x-a)^2+b$ with vertex at $(a, b)$ $(a,b)$

$y = x^{2} + 7 x + 12$ $y=x^2+7x+12$

$\to y = x^{2} + 7 x + {(\frac{7}{2})}^{2} + 12 - {(\frac{7}{2})}^{2}$ $rarr y= x^2+7x+(7/2)^2+12 -(7/2)^2$

$\to y = {(x + \frac{7}{2})}^{2} + (\frac{48}{4} - \frac{49}{4})$ $rarr y = (x+7/2)^2 + (48/4 - 49/4)$

$\to y = {(x - (- \frac{7}{2}))}^{2} + (- \frac{1}{4})$ $rarr y = (x- (-7/2))^2 + (-1/4)$

$XXXXXX$ $color(white)("XXXXXX")$ which is the vertex form with the vertex at $(- \frac{7}{2}, - \frac{1}{4})$ $(-7/2,-1/4)$

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