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

• 是不是 所有抗體都是Y 形，無1例外﹖@1@
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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