Find the range values of c for which 3x^2 - 9x + c > 2.25 for all values of c.
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
of course if you make c bigger the vertex moves up :)
scrap my answer, go with Damon
To find the range of values of c for which the given inequality holds true, we need to solve the inequality for different values of c. Let's start by rearranging the inequality:
3x^2 - 9x + c > 2.25
Subtracting 2.25 from both sides:
3x^2 - 9x + c - 2.25 > 0
Combining like terms:
3x^2 - 9x + (c - 2.25) > 0
Now we have a quadratic inequality. To solve it, we can use the method of factoring or the quadratic formula. However, since we're interested in the range of values for c, we'll focus on analyzing the discriminant of the quadratic equation.
The discriminant of a quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac. In our case, the discriminant is:
D = (-9)^2 - 4(3)(c - 2.25)
= 81 - 12(c - 2.25)
= 81 - 12c + 27
= -12c + 108
For the quadratic inequality to have solutions, the discriminant D must be greater than zero (D > 0). Therefore, we have:
-12c + 108 > 0
Adding 12c to both sides:
108 > 12c
Dividing both sides by 12 (remembering to flip the inequality when dividing by a negative number):
9 > c
So, the range of values for c that satisfy the inequality 3x^2 - 9x + c > 2.25 for all values of c is c < 9.
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