Find the x-intercepts?
929261
y/9 = 250x^3 + 150 x^2 + 55x + 70\n\nTo get the x-intercepts, we set y=0:\n\n250x^3 + 150 x^2 + 55x + 70 = 0\n\n50x^3 + 30x^2 + 11x + 14 = 0\n\nLet f(x) = 50x^3 + 30x^2 + 11x + 14\n\nWe want to see where f(x)=0.  We note that f(0) = 14, and that for all positive x, f(x) > 0, so there are no positive roots.  Also, f(-1) = -50 + 30 - 11 + 14 = -17, so there's a root between -1 and 0.  Also, for all x < -1, f(x) < 0, so there are no roots less than -1.  All the roots are between -1 and 0.\n\nWe have two points, (0,14) and (-1,-17).  Let's try x = -0.5:\n\nf(-0.5) = 9.75  ==>  (-0.5, 9.75)  (I'm sort of interpolating)\n\nf(-0.7) = 3.85  ==>  (-0.7, 3.85)\n\nf(-0.8) = -1.2  ==>  (-0.8,-1.2)\n\nf(-0.78) = -0.06  ==>  (-0.78, -0.06)\n\nThat's pretty good, although we could go to a third decimal if we wanted to.  Just as a check, f(-0.785) = -0.34.  Let's go with one root at x = -0.78.\n\nNow let's divide that root out, reducing the cubic to a quadratic:\n\nf(x) = (x + 0.78)(50x^2 - 9x + 18) = 0\n\nThe roots of the quadratic g(x) = 50x^2 - 9x + 18 = 0 are imaginary.  Using the quadratic formula,\n\nx = 0.01 [9 +/- sqrt(81 - 3600)] = .01 [9 +/- 3 sqrt(9 - 400)]\n\nx = .03 (3 +/- i sqrt 391)\n\nso your only real root is x = -0.78.  (Answer)
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.
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.
431677
If you're dealing with a straight line (and a few other types of line I think), the equation will be of the form\ny=mx+c\nThe c part will give you the y-intercept i.e. where the line crosses the y-axis.
946786
It is notable that the x-intercepts, or solution, of a quadratic equation are known as the ROOTS or the ZEROS of that particular equation.\n\nWhen you solve a quadratic equation, you are looking for the points that make the equation 0; recall that all problems have one side set to zero. For example:\n\nx^2 + 2x + 1 = 0, the original being y=x^2 + 2x + 1\n\nWhat you do when you solve the problem is look for the input values of x that give a y value of 0. It is precisely this that produces the effect of a solution being an x-intercept. Continuing:\n\n(x+1)(x+1) = 0, x = -1\n\nWhat you have done was assume that y=0 (and hence restricted yourself to the x-axis), then tried to find the x points where this \noccurs. By setting y=0, you are "searching" along the x-axis; the ssolution is given where the function CROSSES the axis.
349372
Intercept :  The coordinate of a point at which a line, curve, or surface intersects a coordinate axis.\n So to find X-intercept the line has to intersect the y coordinate axis.therefore put y=0 in the given equation. So\n\n6x-3*0=18\nSo 6x=18\nand So x=18/6\n\nTherefore the result is : x=3\n\nX-intercept is 3
845468
Y-intercepts :   .5\n\nX-intercept:   1\n\nAssymptote at x = 2
819736
x intercept is when y=0\n\nso this would be x=3\n\n3^2 - 3 -6=0\n9-3-6=0\nso x intercept is 3\n\nproof:\n0= x^2 - x - 6\n6 = x ( x-1)\n6 = 3 (3-1)\n6 = 3 * 2\n6=6
887226
x,y intercepts are\n9-y=+/-(x-2)^2\ny intercept bysetting x=0 will be (0,9)\nx intercept by setting y=0\n9=+/-(x-2)^2\nx-2=3x=5\nx-2=-3\nx=-1\nx intercepts(-1,0) and (5,0)\nvertex=(2,9)
304476
y = x^2-5x\n0 = x^2-5x\n\nx^2-5x = 0\n\nx(x-5) = 0\n\nx = 0 or x = 5\n\nx-intercepts : (0,0) and (5,0)\n\nANSWER : two
699032
wx - x =\n\nx(w - 1)\n\n- - - - - - -s-
746165
Remember the fomula...\n\ny = mx + b\n\nm = Slope\n\nb = Y-intercept\n\nx and y = points\n\nSo plug everything in...\n\n1 = 4(2) + b\nMulitply\n1 = 8 + b\nSubtract 8 on each side\n-7 = b\n\ny = 4x - 7
176045
x-intercept\ny-intercept
1169410
An x-intercept occurs where y = 0, so plug in 0 for y.\n\nx^2 + x - 6 = 0\n\nFactor the quadratic equation (reverse FOIL)\n\n(x + 3)(x - 2) = 0\n\nx = 2, -3
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)
704179
An equation in slope-intercept form is y = mx + b\nIn the equation, m = slope and b = y-intercept\n\nUsing the equation given\ny=-3/2x+6\ny = mx + b\n\nWe know that the slope is -3/2.  That would be m.\nWe know the y-intercept is 6.  That would be b.\n\nNow we have to find the x-intercept.\n\nTo get the x-intercept, we will set y = 0 and solve the equation for x.\ny=-3/2x+6\nSubstitute 0 for y\n0=-3/2x+6\nAdd 3/2x to both sides\n3/2x = 6\nMultiply both sides by 2/3\n2/3*3/2x = 6*2/3\nx = 4\nThe x-intercept is 4.\n\nSlope =  -3/2. \ny-intercept = 6.\nx-intercept = 4.
147089
y intercept: (0,4)\nx intercept: (3,0)