# SOLUTION: Given y=9x^2-6x+3, What is the equation of the axis of symmetry? a)x=2/3 b)x=-1/3 c)x=-2/3 d)x=1/3 what are the coordinates of the vertex? Y-intercepts: X-intercepts:

Algebra ->  Algebra  -> Rational-functions -> SOLUTION: Given y=9x^2-6x+3, What is the equation of the axis of symmetry? a)x=2/3 b)x=-1/3 c)x=-2/3 d)x=1/3 what are the coordinates of the vertex? Y-intercepts: X-intercepts:       Log On

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 Question 138225: Given y=9x^2-6x+3, What is the equation of the axis of symmetry? a)x=2/3 b)x=-1/3 c)x=-2/3 d)x=1/3 what are the coordinates of the vertex? Y-intercepts: X-intercepts: I think it opens up, but that is as far as i got... I appreciate any help! Thank you!Answer by solver91311(16877)   (Show Source): You can put this solution on YOUR website!Given , the x-coordinate of the vertex of the parabola is given by , hence the axis of symmetry is . The y-coordinate of the vertex is the value of the function at the x-coordinate of the vertex. So the vertex is at (,) The x-intercepts are the values of x that make the function = 0, so set your function equal to 0 and solve the quadratic equation. If you end up with real roots, the x-intercepts are (x1,0) and (x2,0). If you end up with a conjugate pair of complex roots, then there are no x-intercepts. The y-intercept is the value of the function at 0, so substitute 0 for x and do the arithmetic. The y intercept is (0,f(0)) And you are right, it does open up. A parabola opens upward whenever , and downward when