given f(x) =5x2-3x+1 find f(-2) i say -13?
143790
This is an ambigous way to write the function, there are a couple of answers:\n\nIf it means:\n(5)(2x)-3x+1\nIt equals -13\n\nIf it means:\n(5)(2)-3x+1\nIt equals 17\n\nIf it means:\n5x^2-3x+1\nIt equals 27
874921
-7x-2>=3x-21\nadding -3x+2\n-10x>=-19\n10x<=19\ndividing by10\nx<=1.9
958467
Replace x with 2:\nf(2)=2^3-4(2)^2+14(2)-20\nf(2)=8-16+28-20\nf(2)=-8+8\nf(2)=0
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
838016
This is basically just asking what is f(x)/g(x)\n\nso your answer would be \n\n(x+2) / (3x^2 - 5)
838764
- 7 - 4x < 13\n\n-7 - 4x + 7 < 13 + 7\n\n- 4x < 20\n\n- 4x/- 4 > 20 /- 4\n\nx > - 5\n\nThe answer is x > - 5\n\n- - - - -s-
888899
f(x) = x^2 + 3x\n\nf(2-a) = (2-a)^2 + 3(2-a)\n\nExpand (2-a)^2 to (2-a)(2-a) and distribute the 3(2-a)\n\nf(2-a) = (2-a)(2-a) + 6 - 3a\n\nf(2-a) = 4 -2a -2a + a^2 + 6 - 3a\n\nThe solution is:\n\nf(2-a) = 10 - 7a + a^2
860831
I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)
759378
More concisely, f(-5)=-8, f(-3)=-2, f(-2)=1, f(1)=10.\n\nf(x) can equal anything as long as it fits these points. It seems to form a linear line as shown above...
879979
f(x)=(x^2-2x)(x^2+2.44x+1.4884)\n =(x^4+2.44x^3+1.4884x^2-2x^3-4.88x^2- 2.9768x)\n =(x^4 + 0.44x^3 - 3.3916x^2 - 2.9768x)\n\nF(x)=(x^5)/5+(0.44x^4)/4-(3.3916x^3)/3-             (2.9768x^2)/2+c\n\nam i really asked a Q?
559365
f(x)=x+3\nif x=0, f(0)=0+3=3\nif x=2, f(2)=2+3=5\nif x=-2, f(-2)=-2+3=1\n\nfor g(x)=2x+1\nif x=0, g(0)=2*0+1=1\nif x=2, g(2)=2*2+1=5\nif x=5, g(5)=2*5+1=11\n\nfor h(x.y)=2x+y\nif (x,y)=(0,0) h(0,0)=2*0+0=0\nif (x,y)=(2,1) h(2,1)=2*2+1=5\nif (x,y)=(1,3) h(1,3)=2*1+3=5
889441
Hello dear Neil \n\n5x^2+3x+1=0\n\n● Step 1 \nif F(x) = ax^2 + bx + c \n∆ = b^2 - 4(a)(c) & ∆ ≥ 0\nF(x) = 5x^2+3x+1\na = +5 , b = +3 , c = +1\n∆ = (3^2) - 4 (5)(1) \n∆ = 9 - 20 \n∆ = -11 < 0\nwell this function has so real solution\n\n● Step 2 \nx1 = 1/10 (-3 - i √11)\nx2 = 1/10 (-3 + i √11)\nso \nx 1 = -0.3 - 0.331662i\nx 2 = -0.3 + 0.331662i\n\nSo you are pretty right .\nGood Luck Neil ♣
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even
933970
f-g(x)=(x+2)-(3x^2-5) So\nf-g(x)=x+7-3x^2
1047777
first x outside x inside x last\n(2x*5x) + (2x * -2) + (-5 * 5x) + (-5 * -2)=\n10x^2 + -4x + -25x + 10=\n10x^2 - 29x + 10
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
216211
when x=0 \ny=f(x)=\n(-1/15)(-5)(3)^2(-1)^3=\n(-1/15)(-5)(9)(-1)=\n(5/15) *(-9)=\n(1/3)*(-9)=\n-9/3=-3\n\n(0,-3) y-intercept\n\n\nWhen y=0\n\n-1/15(x-5)(x+3)^2(x-1)^3=0 multiply by -15\n(x-5)(x+3)^2(x-1)^3=0\n\nso x-5=0 and x=5  or\n   x+3=0 and x=-3 or\n   x-1=0 and x=1\n\nso (5,0) (-3,0) and (1,0) are x intercepts.