What is the vertex form of $y = 7 x^{2} + 3 x + 5$ $y=7x^2 +3x + 5$ ?

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What is the vertex form of $y = 7 x^{2} + 3 x + 5$ $y=7x^2 +3x + 5$ ?

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Answer 1 · 2016-01-06T12:14:48

$y = 7 {(x + \frac{3}{14})}^{2} + \frac{917}{196}$ $y = 7(x + 3/14)^2 + 917/196$

Explanation:

The vertex form of a quadratic equation $y = a x^{2} + b x + c$ $y=ax^2 +bx +c$ is $y = a {(x + m)}^{2} + n$ $y=a(x+m)^2 +n$ , where $m = \frac{b}{2 a}$ $m = b/(2a)$ and $n = - a {(\frac{b}{2 a})}^{2} + c$ $n = -a(b/(2a))^2 + c$
Then the vertex is at the point where the bracketed expression is zero and is therefore $(- m, n)$ $(-m,n)$
Therefore $y = 7 {(x + \frac{3}{14})}^{2} - 7 \cdot \frac{9}{196} + 5$ $y = 7(x+3/14)^2 -7*9/196 +5$
$y = 7 {(x + \frac{3}{14})}^{2} - \frac{63 + 980}{196}$ $y = 7(x +3/14)^2 - (63 + 980)/196$
$y = 7 {(x + \frac{3}{14})}^{2} + \frac{917}{196}$ $y = 7(x + 3/14)^2 + 917/196$

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What is the vertex form of $y = 7 x^{2} + 3 x + 5$ $y=7x^2 +3x + 5$ ?

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Explanation:

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