first let's find the discriminate, b2-4ac = (-12)2 - 4(2)(20) = 144 - 160 = -16. The roots will be imaginary, so let's
complete the square:
2[ (x2 - 6x) ] + 20 = y
2[ (x + (-6/2))2 - (-6/2)2 ] + 20 = y 'subt.that 9 that we introduced by compl.sq
2[ (x - 3 )2 - 9 ] + 20 = y
2 (x - 3 )2 - 18 + 20 = y 'distributing the 2 we get the 18
2 (x - 3 )2 + 2 = y
The parabola has its vertex at (3, 2), I think, but let's check that axis of symmetry:
-b 12
xsym = -------- which from the original equation = ------- = 3, so that checks.
2a 2•2
Check to see the turning point is (3,2) by subbing 3 for x in the original:
2(3)2 - 12(3) + 20 =
18 - 36 + 20 = 2 so that checks.
The parabola opens upward because a = +2 > 0.
the y.intercept is 20 (the value of y when x = 0)
the pt.of symmetry with respect to the y.intercept is 3 units to the left of the
axis of symmetry or 3 right of the y.intercept plus 3 more to the right, or
another way to say it is 6 units to the right of the y.intercept at the pt. (6,20).
Using the quadratic formula gives the roots at 3 ± i. That will give two more easy points (think of the projection on the x.axis), or sub 3 ± i into the original equation.
Got about 3 points graphed here - probably the most important ones; did some of it two ways.
The first way, completing the square is Algebra 2, the second way is Algebra 1.
