-3x^2+2x-7

-3x^2+2x-7
=-3[x^2 -(2/3) x] - 7
=-3[x^2 - (2/3)x + (1/3)^2] - 7 + 3(1/3)^2
= -3(x - 1/3)^2 - 20/3
since (x-1/3)^2 >= 0 for all real x, -3(x-1/3)^2 <=0
Hence -3(x - 1/3)^2 - 20/3 < 0
Hence -3x^2+2x-7 < 0 for all real x.

2. Consider 2x^2+2ax-(b+1) = 0 ...(1)
and x^2+ {(b-a)/2}x -(b-1) = 0 ...(2)
(2)x2: 2x^2 + (b-a)x - 2(b-1) = 0
If the two graphs cuts at the same points P and Q,
2a = b - a and b + 1 = 2(b - 1)
b = 3a and b + 1 = 2b - 2
b = 3a and b = 3
a = b/3 = 1, and b = 3
Therefore, a = 1, b = 3

-3x^2+2x-7
=-3(x^2-2/3x+7/3)
=-3(x^2-2/3x+1/3^2+7/3-1/3^2)
=-3(x-1/3)^2-20/3

Therefore the maximum value of fuction is -20/3
Therefore all value of function is negative

y=2x^2+2ax-(b+1)
y= x^2+ {(b-a)/2}x -(b-1)

In this question ,you can consider that y=0 and consider the sum and product of roots of two equation.

-3x^2+2x-7
note that △=(-2)^2-4(-3)(-7)
= -80 &lt; 0 and a &lt; 0 (&quot;a&quot; means -3)
hence, -3x^2+2x-7 is always negative for all real values of x.