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

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Answer 1 · 2017-11-23T12:15:52

Vertex form of equation is $y = 10 {(x + 0.55)}^{2} - 15.025$ $y= 10(x+0.55)^2-15.025$

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

$y = 6 x^{2} + 12 x - 6 x - 12 + 4 x^{2} + 5 x$ $y=6x^2+12x-6x-12+4x^2+5x$ or

$y = 10 x^{2} + 11 x - 12 or y = 10 (x^{2} + \frac{11}{10} x) - 12$ $y=10x^2+11x-12 or y= 10(x^2+11/10x)-12$ or

$y = 10 {x^{2} + \frac{11}{10} x + {(\frac{11}{20})}^{2}} - 10 \cdot {(\frac{11}{20})}^{2} - 12$ $y= 10{x^2+11/10x+(11/20)^2}-10*(11/20)^2-12$ or

$y = 10 {(x + \frac{11}{20})}^{2} - 3.025 - 12$ $y= 10(x+11/20)^2-3.025-12$ or

$y = 10 {(x + 0.55)}^{2} - 15.025$ $y= 10(x+0.55)^2-15.025$ .Comparing with standard vertex form

of equation $f (x) = a {(x - h)}^{2} + k; (h, k)$ $f(x)=a(x-h)^2+k ; (h,k)$ being vertex we find

here $h = - 0.55, k = - 15.025$ $h= -0.55 , k=-15.025$ So vertex is at

$(- 0.55, - 15.025)$ $(-0.55,-15.025)$ and vertex form of equation is

$y = 10 {(x + 0.55)}^{2} - 15.025$ $y= 10(x+0.55)^2-15.025$ [Ans]

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