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what is the axis of symmetry,vertex,and y intercept of the quadratic equation y=x^2-8x+2

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finding the axis of symmetry,vertex,and y intercept of the quadratic equation

asked Dec 5, 2013 in GEOMETRY by homeworkhelp Mentor

1 Answer

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Standard equation of parabola is y = ax^2+bx+c

Compare the given equation y = x^2-8x+2 to above equation.

a = 1, b = -8, c = 2

Axis of symmetry is x = -b/2a

x = -(-8)/2(1)

x = 4

Vertex of parabola

Substitute the x value in equation y = x^2-8x+2

y = (4)^2-8(4)+2

y = 16-32+2

y = -14

Vertex of parabola is (4,-14).

To find y intercept substiute x = 0 in y = x^2-8x+2.

y = (0)^2-8(0)+2

y = 2

y intercept is 2.

answered Dec 13, 2013 by david Expert

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