What is the vertex form of $y = (9 x - 6) (3 x + 2) + 4 x^{2} + 5 x$ $y= (9x-6)(3x+2)+4x^2+5x$ ?

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Answer 1 · 2017-04-25T12:44:42

$y = 31 {(x + \frac{5}{62})}^{2} - \frac{1513}{124}$ $y=31(x+5/62)^2-1513/124$

$y = (9 x - 6) (3 x + 2) + 4 x^{2} + 5 x$ $y=(9x-6)(3x+2)+4x^2+5x$

= $27 x^{2} + 18 x - 18 x - 12 + 4 x^{2} + 5 x$ $27x^2+18x-18x-12+4x^2+5x$

= $31 x^{2} + 5 x - 12$ $31x^2+5x-12$

= $31 (x^{2} + \frac{5}{31} x) - 12$ $31(x^2+5/31x)-12$

= $31 (x^{2} + 2 \times \frac{5}{62} \times x + {(\frac{5}{62})}^{2} - {(\frac{5}{62})}^{2}) - 12$ $31(x^2+2xx5/62xx x+(5/62)^2-(5/62)^2)-12$

= $31 {(x + \frac{5}{62})}^{2} - 31 {(\frac{5}{62})}^{2} - 12$ $31(x+5/62)^2-31(5/62)^2-12$

= $31 {(x + \frac{5}{62})}^{2} - \frac{25}{124} - 12$ $31(x+5/62)^2-25/124-12$

or $y = 31 {(x + \frac{5}{62})}^{2} - 12 \frac{25}{124}$ $y=31(x+5/62)^2-12 25/124$

i.e. $y = 31 {(x + \frac{5}{62})}^{2} - \frac{1513}{124}$ $y=31(x+5/62)^2-1513/124$

and vertex is $(- \frac{5}{62}, - 12 \frac{25}{124})$ $(-5/62,-12 25/124)$

graph{y=31(x+5/62)^2-1513/124 [-3, 3, -20, 20]}

What is the vertex form of y=(9x−6)(3x+2)+4x2+5x y= (9x-6)(3x+2)+4x^2+5x?