貨運百科

標題:

-3x^2+2x-7

發問:

Show that -3x^2+2x-7 is always negative for all real values of x. (5識點寫 d steps) THe graph of the functions y=2x^2+2ax-(b+1) and y= x^2+ {(b-a)/2}x -(b-1) cut the same points P and Q on the x-axis. Find the values of a and b. Plz show the steps.THZ>

最佳解答:

-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

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其他解答:

-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 < 0 and a < 0 ("a" means -3) hence, -3x^2+2x-7 is always negative for all real values of x.