Find the vertex of y=-(x-1)(x+4)?

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Find the vertex of y=-(x-1)(x+4)?

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Answer 1 · 2018-07-09T05:50:02

Vertex $(- \frac{3}{2}, \frac{25}{4})$ $(-3/2, 25/4)$

Explanation:

Given -

$y = - (x - 1) (x + 4)$ $y=-(x-1)(x+4)$

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

$y = - (x^{2} + 3 x - 4)$ $y=-(x^2+3x-4)$

$y = - x^{2} - 3 x + 4$ $y=-x^2-3x+4$

$x = \frac{- b}{2 a} = \frac{- (- 3)}{2 \times - 1} = \frac{3}{- 2} = - \frac{3}{2}$ $x=(-b)/(2a)=(-(-3))/(2 xx -1)=3/(-2)=-3/2$

At $x = - \frac{3}{2}; y = - {(- \frac{3}{2})}^{2} - 3 (- \frac{3}{2}) + 4$ $x=-3/2; y=-(-3/2)^2-3(-3/2)+4$

$y = - \frac{9}{4} + \frac{9}{2} + 4 = \frac{- 9 + 18 + 16}{4} = \frac{25}{4}$ $y=-9/4+9/2+4=(-9+18+16)/4=25/4$

Vertex $(- \frac{3}{2}, \frac{25}{4})$ $(-3/2, 25/4)$

Answer 2 · 2018-07-09T12:06:48

$(\frac{3}{2}, \frac{25}{4})$ $(3/2, 25/4)$

Explanation:

$y = - (x - 1) \cdot (x + 4)$ $y=-(x-1)*(x+4)$

$y = - x^{2} - 3 x + 4$ $y=-x^2-3x+4$

$y = - x^{2} - 3 x - {(\frac{3}{2})}^{2} + 4 + {(\frac{3}{2})}^{2}$ $y=-x^2-3x-(3/2)^2+4+(3/2)^2$

$y = \frac{25}{4} - {(x - \frac{3}{2})}^{2}$ $y=25/4-(x-3/2)^2$

Hence vertex of this equation is $(\frac{3}{2}, \frac{25}{4})$ $(3/2, 25/4)$

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Find the vertex of y=-(x-1)(x+4)?

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