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

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

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Answer 1 · 2015-12-29T12:44:30

$(\frac{7}{3}, - \frac{10}{3})$ $(7/3, -10/3)$

Explanation:

First expand and simplify to get get one term for each power of x.
$y = 4 x^{2} - 12 x + 9 - x^{2} - 2 x + 4$ $y = 4x^2 -12x + 9 - x^2 - 2x + 4$
$y = 3 x^{2} - 14 x + 13$ $y = 3x^2 -14x + 13$
$y = 3 (x^{2} - \frac{14 x}{3} + \frac{13}{3})$ $y = 3(x^2 -(14x)/3 +13/3)$
Use completing the square to put the expression into vertex form
$y = 3 {(x - \frac{7}{3})}^{2} - \frac{49}{9} + \frac{13}{3}) = 3 ({(x - \frac{7}{3})}^{2} - \frac{10}{9})$ $y = 3(x - 7/3)^2 -49/9 + 13/3) = 3((x-7/3)^2 -10/9)$
$y = 3 {(x - \frac{7}{3})}^{2} - \frac{10}{3}$ $y = 3(x-7/3)^2 -10/3$
Then the vertex occurs where the bracketed term is zero.
Vertex is $(\frac{7}{3}, - \frac{10}{3})$ $(7/3, -10/3)$

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

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

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Explanation:

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