What is the vertex form of $y = (x - 14) (3 x + 4) - x^{2} + 2 x$ $y= (x-14)(3x+4)-x^2+2x$ ?

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Answer 1 · 2016-11-06T22:28:17

$y = 2 {(x - 9)}^{2} - 218$ $y=2(x-9)^2-218$

$y = 3 x^{2} + 4 x - 42 x - 56 - x^{2} + 2 x = 2 x^{2} - 36 x - 56$ $y=3x^2+4x-42x-56-x^2+2x=2x^2-36x-56$

If we complete the square with thefirst two terms

$y = 2 {(x - 9)}^{2} - 162 - 56 = 2 {(x - 9)}^{2} - 218$ $y=2(x-9)^2-162-56=2(x-9)^2-218$

What is the vertex form of y=(x−14)(3x+4)−x2+2x y= (x-14)(3x+4)-x^2+2x?