What is the vertex of $y = 2 {(\frac{x}{2} - 2)}^{2} - 4$ $y=2(x/2-2)^2-4$ ?

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What is the vertex of $y = 2 {(\frac{x}{2} - 2)}^{2} - 4$ $y=2(x/2-2)^2-4$ ?

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Answer 1 · 2018-07-10T06:01:51

Vertex $(4, - 4)$ $(4, -4)$

Explanation:

Given -

$y = 2 {(\frac{x}{2} - 2)}^{2} - 4$ $y=2(x/2-2)^2-4$

$y = 2 (\frac{x^{2}}{4} - 2 x + 4) - 4$ $y=2(x^2/4-2x+4)-4$

$y = \frac{1}{2} x^{2} - 4 x + 8 - 4$ $y=1/2 x^2-4x+8-4$

$y = \frac{1}{2} x^{2} - 4 x + 4$ $y=1/2 x^2-4x+4$

Vertex -

$x = \frac{- b}{2 a} = \frac{- (- 4)}{2 \times \frac{1}{2}} = \frac{4}{1} = 4$ $x=(-b)/(2a)=(-(-4))/(2 xx 1/2)=4/1=4$

At $x = 4; y = 2 {(\frac{4}{2} - 2)}^{2} - 4 = 2 (0) - 4 = - 4$ $x=4; y=2(4/2-2)^2-4=2(0)-4=-4$

Vertex $(4, - 4)$ $(4, -4)$

Answer 2 · 2018-07-10T06:20:14

$(4, - 4)$ $(4, -4)$

Explanation:

Given equation,

$y = 2 {(\frac{x}{2} - 2)}^{2} - 4$ $y=2(x/2-2)^2-4$

$y = 2 {(\frac{x - 4}{2})}^{2} - 4$ $y=2(\frac{x-4}{2})^2-4$

$y = \frac{1}{2} {(x - 4)}^{2} - 4$ $y=1/2(x-4)^2-4$

$y + 4 = \frac{1}{2} {(x - 4)}^{2}$ $y+4=1/2(x-4)^2$

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

The above equation shows a vertical parabola with vertex

$(x - 4 = 0, y + 4 = 0)$ $(x-4=0, y+4=0)$

$(x = 4, y = - 4)$ $(x=4, y=-4)$

$(4, - 4)$ $(4, -4)$

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What is the vertex of $y = 2 {(\frac{x}{2} - 2)}^{2} - 4$ $y=2(x/2-2)^2-4$ ?

2 Answers

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What is the vertex of $y = 2 {(\frac{x}{2} - 2)}^{2} - 4$ $y=2(x/2-2)^2-4$ ?