6. Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y = –2x2 +6x – 1. A.x = –1.5; vertex: (–1.5, –5.5) B.x = –1.5; vertex: (–1.5, – 14.5) C.x = 3; vertex: (3, 35) D.x = 1.5; vertex: (1.5, 3.5)

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6. Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y = –2x2 +6x – 1. A.x = –1.5; vertex: (–1.5, –5.5) B.x = –1.5; vertex: (–1.5, – 14.5) C.x = 3; vertex: (3, 35) D.x = 1.5; vertex: (1.5, 3.5)

Mathematics
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this is an algebra question
well.... let us first find the vertex then for a quadratic equation, that is a 2nd degree one the vertex is at \(\bf \textit{vertex of a parabola}\\ \quad \\ y = {\color{red}{ -2}}x^2{\color{blue}{ +6}}x{\color{green}{ -1}}\qquad \left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\) so... see what you get for the vertex since the axis of symmetry, passes by the vertex
it would be D

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oh ok. thanks

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