y = 3 x^2 - 9 x + c is an upward opening parabola so we want the vertex at y = 2.25
3 x^2 - 9 x = y-c
x^2 - 3 x = y/3 - c/3
x^2 - 3 x + 9/4 = y/3 -c/3 + 9/4
(x-3/2)^2 = (1/3)(y -c +27/4)
c-27/4 = 2.25 = 9/4
c = 36/4 = 9
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Now check
if c = 9
3 x^2 - 9 x + 9 = y
x^2 - 3 x = y/3-3
x^2 -3 x + 9/4 = y/3 -12/4 + 9/4
(x-3/2)^2 = y/3 - 3/4 = (1/3 ) (y- 9/4)
9/4 = 2.25 sure eenough
Find the range values of c for which 3x^2 - 9x + c > 2.25 for all values of c.
4 answers
of course if you make c bigger the vertex moves up :)
consider 3x^2 - 9x + c - 225 = 0
and let's look at the discriminant
b^2 - 4ac has to be greater than 0
81 - 4(3)(c-22) > 0
81 - 12c + 264 > 0
-12c > -345
c < 345/12
c < 115/4 c < appr 28.75
and let's look at the discriminant
b^2 - 4ac has to be greater than 0
81 - 4(3)(c-22) > 0
81 - 12c + 264 > 0
-12c > -345
c < 345/12
c < 115/4 c < appr 28.75
scrap my answer, go with Damon