What is the vertex form of $y = x^{2} + 7 x - 3$ $y=x^2+7x-3$ ?

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What is the vertex form of $y = x^{2} + 7 x - 3$ $y=x^2+7x-3$ ?

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Answer 1 · 2016-01-14T15:58:49

$y = {(x + \frac{7}{2})}^{2} - \frac{61}{4}$ $y= (x+7/2)^2 - 61/4$ or $4 y = {(2 x + 7)}^{2} - 61$ $4y = (2x+7)^2 -61$

Explanation:

For a quadratic of the form $y = a x^{2} + b x + c$ $y=ax^2 +bx +c$ the vertex form is $y = a [{(x + \frac{b}{2 a})}^{2} - {(\frac{b}{2 a})}^{2}] + c$ $y = a[(x+b/(2a))^2 -(b/(2a))^2] +c$
In this case that gives us
$y = {(x + \frac{7}{2})}^{2} - \frac{49}{4} - 3$ $y = (x+7/2)^2 - 49/4 - 3$
$y = {(x + \frac{7}{2})}^{2} - \frac{61}{4}$ $y= (x+7/2)^2 - 61/4$

The vertex is then $(- \frac{7}{2}, - \frac{61}{4})$ $(-7/2, -61/4)$

Multiplying throughout by $4$ $4$ gives
$4 y = {(2 x + 7)}^{2} - 61$ $4y = (2x+7)^2 -61$

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What is the vertex form of $y = x^{2} + 7 x - 3$ $y=x^2+7x-3$ ?

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What is the vertex form of $y = x^{2} + 7 x - 3$ $y=x^2+7x-3$ ?