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stevetron
 one year ago
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)
stevetron
 one year ago
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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stevetron
 one year ago
Best ResponseYou've already chosen the best response.0this is an algebra question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well.... 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
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