Determine the left-hand behavior and the right-hand behaior of the graph of -3/4(x^3+2x^2-3x+1)?
824215
Did you mean -(3/4) * (x^3+2x^2-3x+1) ?\n\nLet f(x) =  -(3/4) * (x^3+2x^2-3x+1). Equivalently, f(x) = -(3/4)x^3 - (3/2)x^2 + (9/4)x - 3/4. Consider an arbitrary real valued x say x = c. We need to determine if the left-hand behavior and the right-hand behavior of the function f at x = c are the same or not. \n\nSince f is a polynomial function (of degree 3) on x, it follows that this is continuous on the set of real numbers. Thus, by applying the limit on both sides (left and right) of x, we will obtain the same value for both sides. \n\nThat is, the left-hand Limit will be as follows: as x approaches to the left of c, we make f(x) approaches f(c) = -(3/4)c^3 - (3/2)c^2 + (9/4)c - 3/4. Similarly, the right-hand limit will be: as x approaches to the right c, we also get f(x) approaches to f(c). \n\nTherefore, both the left and right-hand limit yields to f(c); so as x approaches c in any direction, we always obtain f(c). Hence, both left and right-hand behavior of f are the same for any choice of a real valued x.
801068
-4(-1)^2 - 3(-1)\n-4(1) - -3\n-4 + 3\n= - 1
838358
i can't give you the answer but i used my ti-83 and i can tell you the points of the graph...\n(1,0) (2,4) (3,8) (4,0)\n\nHope this helps
221826
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
374365
it will be choice (c)\n(x+1)(x-1)\nx^2+x-x-1\nx^2-1
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_
179915
There will be a vertical asymptote at x=0. When x values get increasingly closer to zero from the right, the graph approaches infinity. When the x values approach 0 from the left, the graph approaches negative infinity.\n\nSimilarly, there is also a horizontal asymptote at y=4. As x approaches infinity, the graph approaches 4 (1/infinity + 4 = 4). Also, as x approaches negative infinity, the graph also approaches 4 (1/negative infinitive + 4 = 4).
445985
if by this you mean\n\n((3x^-2 * y)^-2)/((4xy^-2)^-1)\n\nthis becomes\n\n((3^-2) * x^(-2 * -2) * y^-2)/((4^-1) * x^-1 * y^(-2 * -1))\n\n((1/9)x^4 * y^-2)/((1/4) * x^-1 * y^2)\n\n((1/9)/(1/4)) * x^(4 - (-1)) * y^(-2 - 2)\n\n((1/9)*(4/1)) * x^(4 + 1) * y^(-2 + (-2))\n\n(4/9)x^5y^-4\n\nwithout using negative exponents, the answer would be\n\n(4x^5)/(9y^4)
91768
Quadratic Functions have the form y = ax^2 + bx + c\n\nThe vertex will always have x = -b/(2a). You can calculate this number and plug it into your equation to solve for y. If a is positive, the graph will go up in both directions from the vertex. if a is negative it will go down. You can pick some x value different from the vertex to help you plot other points to allow a symmetric parabola graph.\n\nGood Luck!
876214
g(x)=-x^3+4x-2\ng'(x)=3x^2+4\ng"(x)=6x\ng"(x)=0=6x\nx=0 is the inflection point
791991
Given a(x^2) + bx + c as the quadratic the c determines the vertical shift of the graph. If c = 0 you get a parabola whose vertex touches (0,0). If c > 0 the graph shifts up to have a y-intercept of (0,c). If c <0 you get the same result but (0,c) falls below the x-axis.
986609
y=(x+3)^2=x^2+6x+9\ny'=2x+6=0\n2x=-6\nx=-3\ny(-3+3)^2=0\nso the vertex is (-3,0)
864521
y = 3/(1 - x)²\n\n1, (1 - x)² ≥ 0 for all x and therefore 3/(1 - x)² ≥ 0 for all x\n\n2. 3/(1 - x)² is undefined when x = 0. Therefore 3/(1 - x)² (ie y) → +∞ when x → 1 from the left of 1 and when x → 1 from the right of 1\n\n3. As x → ±∞, 3/(1 - x)² (ie y) → 0\n\n4. Plot a few other points and Bob's your uncle!!
194433
If the number in front of the x^2 term is negative, the graph (a parabola) opens downward, so the vertex would be a maximum.  If it's positive, the parabola opens upward so the vertex is a minimum.\n\nThe vertex is always on the axis of symmetry, so find that first.  The formula for the axis of symmetry of a parabola whose equation is y = ax^2 + bx + c is:\n\nx = -b/(2a)\n\nIf you are familiar with the quadratic equation, this is that equation without the + or - square root part.\n\nTake your first equation:  b = 0 and a = -4, so the equation of the axis of symmetry is x = -0/-8, or x = 0.\n\nFor the second, it is x = -5/2  (since a = 1).\n\nFinally, since the vertex is on the axis of symmetry, it has the same x-coordinate.  So plug its value in for the  x-coordinate of each equation and see what y is.\n\nFirst: y = -4 times 0 + 3 = 3 so vertex = (0,3)\nSecond: y = (-5/2)^2 + 5(5/2) - 12 = 27/4 so vertex = (-5/2, 27/4).\n\nHope that makes sense.
883736
x=-t/2\n2x=-t\n-2x=t\ny=t^2\ny=(-2x)^2\ny=4x^2\nThe graph is a parabola with vertex at (0,0), and in the x-interval [0,5], the end falls at (5, 100).
838574
Part 2 is easy.  The function y = 1/(1+x) has an infinite discontinuity at x = -1.  Also, y(0) = 1, and for large x (positive and negative), the function approaches the x-axis as an asymptote.  For positive x, y(x) > 0, and for large negative x, y(x) < 0.  It's easy to draw the graph.\n\nIn this problem, we care about y(-3) = -1/2; y(-2) = -1; and y(1) = 1/2.  We want the parabola y = ax^2 + bx + c that passes through the points (-3, -1/2), (-2, -1), and (1, 1/2).  Since all three points must satisfy the quadratic equation, we can plug in values and write these three equations:\n\n(1)  For x = -3:  9a - 3b + c = -1/2\n(2)  For x = -2:  4a - 2b + c = -1\n(3)  For x = 1:  a + b + c = 1/2\n\nThe solution to this set of three linear equations in three unknowns is a = 1/4; b = 3/4; and c = -1/2.  (Details of the solution omitted.)\n\nThe quadratic is y = 1/4 x^2 + 3/4 x - 1/2.  This function passes through the three points we identified, and\n\n1/(1+x) < 1/4 x^2 + 3/4 x - 1/2 over the range specified in the problem.  (Actually, there's equality at three points.)\n\nThat inequality can also be written as\n\n4/(1+x) < x^2 + 3x - 2\n\nI'll answer the other part of the question as best I can in the morning.  Right now, it's time to sleep.
843943
4, 0, -2, -2, 0\n\nFigure it out! Don't be lazy!