What is the vertex of $y= 2(x - 4)^2 - 8x+3$ ?

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Answer 1 · 2017-04-26T16:53:07

The vertex is $(6,-27)$

Given: $y= 2(x - 4)^2 - 8x+3$

Expand the square:

$y= 2(x^2 - 8x+16) - 8x+3$

Distribute the 2:

$y= 2x^2 - 16x+32 - 8x+3$

Combine like terms:

$y= 2x^2 - 24x+35$

The x coordinate of the vertex, h, can be computed using the following equation:

$h = -b/(2a)$ where $b = -24$ and $a = 2$

$h = -(-24)/(2(2)$

$h = 6$

The y coordinate of the vertex, k, can be computed by evaluating the function at the value of h, (6):

$k= 2(6 - 4)^2 - 8(6)+3$

$k = -37$

The vertex is $(6,-27)$

What is the vertex of y= 2(x - 4)^2 - 8x+3 y= 2(x - 4)^2 - 8x+3 ?