Inactive Tutor answered 08/05/13
x2 - 2x + 3 = (x - 1)2 + 2
= (x - 1)2 - (-2)
= (x - 1 + sqrt(-2)) (x - 1 - sqrt(-2))
= (x - 1 + sqrt(2) * i) (x - 1 - sqrt(2) * i)
Timothy S.
asked 08/05/13answer for math questions about parabolas
Inactive Tutor answered 08/05/13
x2 - 2x + 3 = (x - 1)2 + 2
= (x - 1)2 - (-2)
= (x - 1 + sqrt(-2)) (x - 1 - sqrt(-2))
= (x - 1 + sqrt(2) * i) (x - 1 - sqrt(2) * i)
Inactive Tutor answered 08/05/13
Parabolas are of the form ax2+bx+c, where a,v and c are the coefficients (numbers) in front of the variable. Notice that all parabolas are of degree 2 (that means the highest power of the variable is 2).
There are three major methods of solving for the roots of a parabola. (roots being the places where the y value will be zero).
1. factor the tri-nomial (three terms) into two binomials (two terms).
2. complete the square (see below)
3. or use the quadratic formula -b + sqrt(b2 - 4ac) then divide the whole thing by 2a
for the parabola stated above: y =
x2 - 2x + 3 factoing will not be the best way.
completing the square is done as follows:
set the equation = 0. 0 = x2 -2x + 3
take the last term (3) and subtract it from both sides which gives you x2 - 2x = -3
take the coefficient of the 'x' term which is -2, take half of it (-1) square it (1)
add this term (1) to both sides which gives you x2 - 2x +1 = -2
NOTICE: you have made a perfect square of the left side which is (x-1)2 so....
(x-1)2 =-2. Now take the sqrt of both sides which gives you x-1 =+sqrt(-2)
so..... the roots are complex! not real. they are x = 1-i*sqrt(2) and 1+i*sqrt(2)
____________________________________________________________________
NOTE: The vertex or (maximum/minimum point) of the parabola ca be gotten by using the co-ordinate pair (-b/2a,f(-b/2a))
((-2/2),6) or ((-1,6)
____________________________________________
Using the quadratic formula we get -(-2) + sqrt(-22-4(a)(3) all divided by (2*1)
which gives you 2 + sqrt(4 - 12) all divided by 2
which gives you 2 + sqrt(-8) all divided by 2, which yields 2 + 2i* sqrt(2) alld divided by 2
whihc is 1 + i*sqrt(2) and 1 - i*sqrt(2)
same answer as completing the square!
Inactive Tutor answered 08/05/13
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.
Inactive Tutor
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.
08/06/13
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.
Inactive Tutor
Wrong solution Grigori. Your factoring works for x2 - 2x - 3.
08/05/13