What is the vertex form of #3y=8x^2 + 17x - 13?

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What is the vertex form of #3y=8x^2 + 17x - 13?

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Answer 1 · 2015-11-25T03:09:37

The vertex form is $y = \frac{8}{3} {(x + \frac{17}{16})}^{2} - \frac{235}{32}$ $y=8/3(x+17/16)^2-235/32$ .

Explanation:

First, let's rewrite the equation so the numbers are all on one side:

$3 y = 8 x^{2} + 17 x - 13$ $3y=8x^2+17x-13$
$y = \frac{8 x^{2}}{3} + \frac{17 x}{3} - \frac{13}{3}$ $y=(8x^2)/3+(17x)/3-13/3$

To find the vertex form of the equation, we must complete the square:

$y = \frac{8 x^{2}}{3} + \frac{17 x}{3} - \frac{13}{3}$ $y=(8x^2)/3+(17x)/3-13/3$

$y = \frac{8}{3} (x^{2} + \frac{17}{8} x) - \frac{13}{3}$ $y=8/3(x^2+17/8x)-13/3$

$y = \frac{8}{3} (x^{2} + \frac{17}{8} x + {(\frac{17}{8} \div 2)}^{2} - {(\frac{17}{8} \div 2)}^{2}) - \frac{13}{3}$ $y=8/3(x^2+17/8x+(17/8-:2)^2-(17/8-:2)^2)-13/3$

$y = \frac{8}{3} (x^{2} + \frac{17}{8} x + {(\frac{17}{8} \cdot \frac{1}{2})}^{2} - {(\frac{17}{8} \cdot \frac{1}{2})}^{2}) - \frac{13}{3}$ $y=8/3(x^2+17/8x+(17/8*1/2)^2-(17/8*1/2)^2)-13/3$

$y = \frac{8}{3} (x^{2} + \frac{17}{8} x + {(\frac{17}{16})}^{2} - {(\frac{17}{16})}^{2}) - \frac{13}{3}$ $y=8/3(x^2+17/8x+(17/16)^2-(17/16)^2)-13/3$

$y = \frac{8}{3} (x^{2} + \frac{17}{8} x + (\frac{289}{256}) - (\frac{289}{256})) - \frac{13}{3}$ $y=8/3(x^2+17/8x+(289/256)-(289/256))-13/3$

$y = \frac{8}{3} (x^{2} + \frac{17}{8} x + (\frac{289}{256})) - \frac{13}{3} - (\frac{289}{256} \cdot \frac{8}{3})$ $y=8/3(x^2+17/8x+(289/256))-13/3-(289/256*8/3)$

$y = \frac{8}{3} {(x + \frac{17}{16})}^{2} - \frac{13}{3} - \frac{289}{96}$ $y=8/3(x+17/16)^2-13/3-289/96$

$y = \frac{8}{3} {(x + \frac{17}{16})}^{2} - \frac{235}{32}$ $y=8/3(x+17/16)^2-235/32$

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What is the vertex form of #3y=8x^2 + 17x - 13?

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