How do you find the vertex of $y = 2 x^{2} - 4 x - 2$ $y=2x^2-4x-2$ ?

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How do you find the vertex of $y = 2 x^{2} - 4 x - 2$ $y=2x^2-4x-2$ ?

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Answer 1 · 2016-01-02T11:10:32

Co-Ordinate of Vertex is (1,-4)

Explanation:

This is equation of a Parabola , the general form of which is $a x^{2} + b x + c$ $ax^2+bx+c$ . So here $a = 2; b = - 4; c = - 2$ $a=2 ; b=-4 ; c=-2$ Let the Co-ordinate of Vertex be $x_{1}, y_{1}$ $x_1,y_1$ . Now we know $x_{1} = - \frac{b}{2 \cdot a}$ $x_1 = -b/(2*a)$
$:.$ $x_1 = 4/4 =1$ Now plugging in $x_1$ in the equation we get $y_1=2*1-4*1-2$ or $y_1 = -4$ So The Co-Ordinate of Vertex
is (1,-4) [Answer]

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How do you find the vertex of $y = 2 x^{2} - 4 x - 2$ $y=2x^2-4x-2$ ?

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How do you find the vertex of y=2x^2-4x-2y=2x2−4x−2 y=2x^2-4x-2?

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How do you find the vertex of $y = 2 x^{2} - 4 x - 2$ $y=2x^2-4x-2$ ?