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

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

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Answer 1 · 2017-07-09T07:33:15

$(- 3, 1)$ $(-3,1)$

Explanation:

Firstly, expand the squared brackets:
$y = 4 (x^{2} + 4 x + 4) - 2 x^{2} - 4 x + 3$ $y=4(x^2+4x+4)-2x^2-4x+3$
Then, expand the brackets:
$y = 4 x^{2} + 16 x + 16 - 2 x^{2} - 4 x + 3$ $y=4x^2+16x+16-2x^2-4x+3$
Collect like terms:
$y = 2 x^{2} + 12 x + 19$ $y=2x^2+12x+19$
Use formula for x-turning point: ( $- \frac{b}{2 a}$ $-b/{2a}$ )
thus, x= $- 3$ $-3$
Plug -3 back into the original formula for y coordinate:
$4 {(- 3 + 2)}^{2} - 2 {(- 3)}^{2} - 4 (- 3) + 3 = 4 - 18 + 12 + 3 = 1$ $4(-3+2)^2-2(-3)^2-4(-3)+3=4-18+12+3=1$
therefore the vertex is: $(- 3, 1)$ $(-3,1)$

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

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Explanation:

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Science

Math

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

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Explanation:

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