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

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Answer 1 · 2016-09-20T02:43:02

Convert to standard form, which is $y = ax^2 + bx + c, a !=0$ .

$y = 2(x - 3)^2 - x + 3$

$y = 2(x^2- 6x + 9) - x + 3$

$y = 2x^2 - 12x + 18 - x + 3$

$y = 2x^2 - 13x + 21$

Now, to determine the vertex, convert to vertex form, which is $y = a(x - p)^2 + q, a !=0$

$y = 2(x^2 - 13/2x + m - m)^2 + 21$

The goal here is to convert to a perfect square. $m$ is given by $(b/2)^2$ , where #b = (ax^2 + bx + ...) inside the parentheses.

$m = ((-13/2)/2)^2 = 169/16$

$y = 2(x^2 - 13/2x + 169/16 - 169/16) + 21$

$y = 2(x^2 - 13/2x+ 169/16) - 169/8 + 21$

$y = 2(x- 13/4)^2 - 1/8$

In vertex form, $y = a(x - p)^2 + q, a !=0$ , the vertex is located at $(p, q)$ . Hence, the vertex is at the coordinates $(13/4, -1/8)$ .

Hopefully this helps!

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