What is the the vertex of $y = x^{2} + 7 x + 12$ $y=x^2+7x +12$ ?

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What is the the vertex of $y = x^{2} + 7 x + 12$ $y=x^2+7x +12$ ?

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Answer 1 · 2016-01-29T07:44:42

$(- \frac{7}{2}, - \frac{1}{4})$ $(-7/2,-1/4)$

Explanation:

Re-express in vertex form by completing the square:

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

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

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

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

The equation:

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

is in vertex form:

$y = a {(x - h)}^{2} + k$ $y = a(x-h)^2+k$

with multiplier $a = 1$ $a=1$ and vertex $(h, k) = (- \frac{7}{2}, - \frac{1}{4})$ $(h,k) = (-7/2,-1/4)$

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What is the the vertex of $y = x^{2} + 7 x + 12$ $y=x^2+7x +12$ ?

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Explanation:

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