About quadratic equation

1.a. f(x) = 2x^2 - kx + 12

= 2(x^2 - kx/2 + (k/4)^2) - k^2/8 + 12

= 2(x - k/4)^2 + 12 - (k^2/8)

For x = 5/2, f(x) attains the minimum

This happens when (x - k/4)^2 = 0

So, 5/2 - k/4 = 0

k = 10

b. For the greatest value of 4 - f(x), f(x) is at minimum value

min f(x) = 12 - (10)^2/8 = -1/2

So, the maximum value of 4 - f(x) = 4 - (-1/2) = 9/2

2.a. 2x^2 + 6x + 9 = a(x + b)^2 + c

= a(x^2 + 2bx + b^2) + c

= ax^2 + 2abx + (ab^2 + c)

Comparing coefficients, a = 2, 2ab = 6, so, b = 6/(2 X 2) = 3/2

(2)(3/2)^2 + c = 9

c = 9/2

b. 2x^2 + 6x + 9 = 2(x + 3/2)^2 + 9/2

The minimum value of this function is 9/2.

c. As minimum value of the equation 2x^2 + 6x + 9 = 9/2

So, 2/(2x^2 + 6x + 9) attains maximum when 2x^2 + 6x + 9 = 9/2

So, 2/(2x^2 + 6x + 9) <= 2/(9/2) = 4/9

And the denominator will converge to +ve infinity as x tends to infinity, so

2/(2x^2 + 6x + 9) > 0

Therefore,

0 < 2/(2x^2 + 6x + 9) <= 4/9