Find the x-and y-intercepts of the rational function?
920959
r(x)= x^3+8/x^2+4\n\nAssuming you mean r(x)= (x^3 + 8)/(x^2 + 4)\n= (x + 2)(x^2 - 2x + 4)/(x^2 + 4)\n\nWhen x = 0, r(x) = 8/4 = 2.\nSo it cuts the y- axis at (0, 2)\n\nr(x) = 0 when (x + 2)(x^2 - 2x + 4)/(x^2 + 4) = 0, ie when x = -2\n\nSo it cuts the x-axis at (-2, 0)\n\nie x-intercept is -2 and y-intercept is 2.
778985
The rational zeros will be of the form p/q where p is a factor of 3 (the lowest order term) and q is a factor of 2 (the coefficient of the highest order x term, in this case 2x^3).  That means the possible answers are +/- 3/1, 3/2, 1/1 and 1/2.  Just plug those in to see which works.  (1, -1, 1.5)
930634
x-intercept is where the curve crosses the x-axis\n\nequation of x-axis is y=0\nseeking the zeroes of x^2-7x+7\nthe values of x that make x^2-7x+7 zero.\n\n\nx^2-7x+7 = 0\nx = (7+/- sqrt(7^2-4.1.7))/2\n= (7 +/- sqrt(21))/2\n\nso the x-intercepts are\n(7+sqrt(21))/2,0) and (7-sqrt(21))/2,0)
885928
Find the X intercept by setting Y = 0\n\nx-2 = 0\nx = 2\n\nFind the Y intercept by setting X = 0\n\ny = 0-2\ny = -2\n\nSo the two points to plot are (0,-2) and (2,0) \nThen connect those points with a line
845468
Y-intercepts :   .5\n\nX-intercept:   1\n\nAssymptote at x = 2
740067
Let's start from the beginning with a quadratic.\n\nLet y=f(x)=ax^2+bx+c, where a is not equal to zero.  The y-intercept is the point where the parabola intersects the y-axis.  So, the point is (0,f(0))=(0,c).\n\nIf the parabola opens upward (a positive) and its vertex is above the x-axis or opens downward (a negative) and its vertex is below the x-axis, there aren't any (real) x-intercepts.  \n\nOtherwise, if the quadratic has x-intercepts, solve ax^2+bx+c=0.  \n\nTry factoring first and if that doesn't work easily, then try using the quadratic formula,\n\nx=(-b &#92;pm &#92;sqrt{b^2-4ac})/(2a).\n\nIf a parabolar has x-intercepts it may have one or two so be sure to check your work carefully.\n\nExamples:\n\ny=x^2+1 has y-intecept (0,1) by no x-intercepts\n\ny=x^2-2x+1=(x-1)^2 has y-intercept (0,1) and x-intercept at (1,0)\n\ny=x^2-x-2=(x-2)(x+1) has y-intercept (0,-2) and x-intercepts (2,0) and (-1,0).\n\nA good idea:  Given a quadratic either try to factor it so it looks like y=(a x+b)(c x+d) or try to write in the form y=a(x-b)^2+c.  In the second case, (b,c) is the vertex and the sign of a tells you whether it goes up or down.  The first case shows that the x-intercepts are x=-b/a and x=-d/c.\n\nExcellent question! :)
698387
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
162736
x-y=5\nx/5-y/5=1\nhence x intercept [5,0] and y intercept [0,-5]
618527
i'm not gonna solve it...but tell u how to solve it.\n\n-x intercept of vertex : -b/2a\n-y intercept of vertex : f(-b/2a)\n-for direction of opening find f(0)\n-i guess the axis of symmetry is the middle point of something. i cant remember that something.\n-find the relationship urself
418521
The Y-INTERCEPT of the line y = b is b.\n\nFor a y-intercept, you plug in 0 for the other variable, primarily x, but in this case, b.\n\nTherefore, if you plug it in, you receive:\n\ny = 0\n0 = 0\n\nThe y-intercept is 0\n\nPerhaps you should check by graphing, but I assure you that I am correct
831482
We just started this stuff in school.  first, think of f(x) as y.  then make a table with the left column as x, and the right column as y.  the vertex is going to be -b/2a.  so in this case.  it'll be -1/4.  that's the middle number in the table and also the vertex.  now put above that number in the table on the x column as so.\nx  y\n1\n0\n-1/4\n-2.5\n-3.5\nnow fill in the the x column by plugging in the x variables.  plug in 0 for x and thats your y-intercept.  then plug in your 0 for y, and thats your x-intercept.  this is going to form a parabola so the axis of symmetry is -1/4.  the domain is going to be all real numbers because there are no contradictory numbers to fill in.  Im not sure how to do the range because im only a junior in high school but I hope I helped a little.
909578
Since the coefficient of x^4 is one, the possible rational zeros are just the factors of 8.\nList all the factors of 8, both the negative and positive.\nAnswer: -1, +1, -2, +2, -4, +4, -8, +8
954948
a) Algebraically the x-coordinate of the vertex is x= -b/(2a)\n\nHere that is x = -10/(2 X 1) = -10/2 = -5\n\nPlug that in for x to get the y-coordinate of the vertex.\n\nb) To get the x-intercepts, replace y (here, g(x)) with 0 and solve.\n\n0 = x^2 + 10x + 21 = (x+7)(x+3)\n\nx = -7 or -3 so there are two x-intercepts: -7 and -3.\n\nc) to find the y-intercepts, replace x with zero. \nSo y = 0 + 0 + 21 = 21  \n21 is the y-intercept.
373363
The y-intercept is where X=0\n\nAny algebraic equation makes this process simple\n\nFor Example.\n\ny=982372987239487x+z\n\nthe y-intercept is always equal to z no matter what interger is place as a factor of x
913484
are you talking about a polynomial :\na + bx +cx^2 + dx^3 +...\n\nbecause those a,b,c,d... do not appear any where on the graph... you have to look at the intercepts and the extreama, and the inflection points to get any good idea about the shape of the graph
1025113
An instructor always wants to know the following about a graph:\n\nThe x-intercepts, the y-intercepts, and for parabolas, where is the vertex.\n\nSet x=0 to find the y intercept.  Hmm (0,0)\n\nSet y=0 and solve to get the x-intercepts:\nIt factors into x(4-x) so, y=0 when x=0 and 4.\n\nYou can always get the vertex by using x= -b/(2a) or, if there are x-intercepts, it will always be halfway in between them.  In this case, the vertex is at x=2.  Plug that in to get y=4.\n\nSo, the important info is:  It opens downward.  The vertext is at (2,4).  The x-intercepts are at (0,0) and (4,0) and the y-intercept is at (0,0).
803558
1.\nc=(xy^2-1)/y\ndifrentiate w.r.t x\n(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0\ny^3 + 2xy^2p - pxy^2 + p=0\ny^3 +xy^2p +p=0\nwhere p=(dy/dx)\n2.\nsimilarly diffrentiate \n(x+y^2)/x^2=-c\nto get\n2(x^2)yp - x^2 - 2xy^2=0\nthanks