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

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

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Answer 1 · 2016-01-10T12:03:20

$V e r t e x (\frac{3}{2}, - \frac{1}{4})$ $Vertex (3/2, -1/4)$

Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{3}{2}$ $x = -b/(2a) = 3/2$
y-coordinate of vertex:
$y (\frac{3}{2}) = (\frac{9}{4}) - 3 (\frac{3}{2}) + 2 = \frac{9}{4} - \frac{9}{2} + 2 = - \frac{9}{4} + \frac{8}{4} = - \frac{1}{4}$ $y(3/2) = (9/4) - 3(3/2) + 2 = 9/4 - 9/2 + 2 = -9/4 + 8/4 = -1/4$
$V e r t e x (\frac{3}{2}, - \frac{1}{4})$ $Vertex (3/2, -1/4)$

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

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

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