# SOLUTION: I need some help badley, find the vertex, the x-intercepts, and the y-intercept of the quadratic function.(remember that the x-coordinate of the vertex is -b/2a) problem y=x2-2x-3

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: I need some help badley, find the vertex, the x-intercepts, and the y-intercept of the quadratic function.(remember that the x-coordinate of the vertex is -b/2a) problem y=x2-2x-3       Log On

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 Question 88210: I need some help badley, find the vertex, the x-intercepts, and the y-intercept of the quadratic function.(remember that the x-coordinate of the vertex is -b/2a) problem y=x2-2x-3 then it says Vertex_________ x-intercepts__________and __________ y-intercept______________ thanks for your helpAnswer by ankor@dixie-net.com(15746)   (Show Source): You can put this solution on YOUR website!I need some help badly, find the vertex, the x-intercepts, and the y-intercept of the quadratic function.(remember that the x-coordinate of the vertex is -b/(2a) problem y=x2-2x-3 then it says : y = x^2 - 2x -3 : Vertex_________ Find the axis of symmetry using x = -b/(2a), in this equation: a=1, b=-2 x = -(-2)/(2*1) x = +2/2 x = +1; Find the vertex by finding y: substitute 1 for x in the original equation: y = 2^1 + -2(1) - 3 y = 1 - 2 - 3 y = -4 Vertex coordinates: x=1, y=-4 : x-intercepts__________and __________ x intercepts occur when y = 0 x^2 - 2x - 3 = 0 This will factor to: (x-3)(x+1) = 0 x = +3 and x = -1 : y-intercept______________ y intercept occurs when x = 0 Substitute 0 for x in the original equation y = 0^2 + 2(0) -3 y = -3 : Did this help you?? Any questions?