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y=x^2-2x+3?

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1 Answer

Where are the qustions? Your parabola is opened up because the first coefficient is positive, The parabola has two roots defined by the equation

                                         x2 -2x + 3 = 0

Factor it:

                                     x2 -2x +3 = (x-3)(x+1) = 0

Thus the roots are   x1 = 3   and x2 = -1. The  axis of symmetry for the parabola is the midpoint of the segment [-1,3], which means x - coordinate of the vertex is  -1 + [3-(-1)]/2 =  1.  Because the parabola is opened up it has only minimum which is equal to y(1) = 2.

Comments

Wrong solution Grigori. Your factoring works for x2 - 2x - 3.

You are right. I messed up with the sign, being in hurry and didn't pay attention on it. The equation doesn't have a real solutions. Quadratic formula gives us:

                           x1 = 1 + iv2,  x2 = 1 - iv2

because the discriminanat is negative. There is no intersection with x -axis. The line of symmetry for the parabola is still the same: x = -b/2a, where b = -2 and a = 1. I apologize for accidental inaccuracy. Thank you for pointing to my miscalcultions.

                          

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