What is the vertex of $y = - 4 x^{2} + 8 x - 19$ $y= −4x^2 + 8x − 19$ ?

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What is the vertex of $y = - 4 x^{2} + 8 x - 19$ $y= −4x^2 + 8x − 19$ ?

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Answer 1 · 2015-12-07T11:13:42

Vertex $(1, - 15)$ $(1, -15)$

Explanation:

Given-

$y = - 4 x^{2} + 8 x - 19$ $y=-4x^2+8x-19$

$x = \frac{- b}{2 a} = \frac{- 8}{2 \times (- 4)} = \frac{- 8}{- 8} = 1$ $x=(-b)/(2a)=(-8)/(2 xx(-4))=(-8)/(-8)=1$

At $x = 1$ $x = 1$

$y = - 4 (1) + 8 (1) - 19 = - 4 + 8 - 19 = - 15$ $y=-4(1)+8(1)-19=-4+8-19=-15$

Vertex $(1, - 15)$ $(1, -15)$

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What is the vertex of $y = - 4 x^{2} + 8 x - 19$ $y= −4x^2 + 8x − 19$ ?

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