What is the coordinates and the vertex of the equasion y=5x^2+5x-2? (^2 equals exponet or squared)?
364111
take the first derivative and set it equal to zero and solve.  2x +1 = 0 x= -1/2  solve for y.  y=-13/4\nvertex is at x= -13/4 y=-2
512448
x^2 + y^2 = 25\nx - y = 0\n\nx = y\n\nY^2 + Y^2 = 25\n2 Y^2 = 25\n\nY ^2 = 25/2\ny =  (25/2)^1/2 or\ny = Sqrt (12.5)
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.
826601
Find where f '(x) = 0 to determine where the vertex is.\n\nf(x) = -3x² + 6x - 2\n\nf′(x) = -6x + 6\n\nf′(x) is 0 when x = 1, so our vertex is at x=1\n\nNow find f(1) to determine the y-value of the vertex:\n\nf(1) = -3(1²) + 6(1) - 2\n= -3 +6 - 2 = 1\n\nSo the vertex is at the point (1,1)
1105592
y+dy=5(x+dx)^2 +1\n  y+dy=5(x² +2xdx +dx²) +1\n    y+dy=5x²+1 +5(2xdx+dx²)\n       dy=10 xdx as dx² is very small\n          dy/dx= 10x
1057389
Look at the coefficient of the x^2 term.  If it is positive, the parabola opens up and has a minimum value; if it is negative, the parabola opens down and has a maximum.  You have 2x^2, so you will have a minimum value.\n\nNext find the vertex.  The x-coordinate of the vertex lies at -b/2a.  So -8/(2)(2) = -8/4 = -2.  Plug this value in to get the y-coordinate of the vertex.  2(-2^2) + (8)(-2) + 5 = 8 - 16 + 5 = -3.\n\nSo you have a minimum value of -3.
133564
The coordinates of the vertex of a parabola \ny = ax^2 + bx + c is\n[-b/2a , (4ac - b^2)/4a]\nJust substitute and you will get\nV(2 , -7)
871684
The graph of this equation looks like a V. It's the same as the graph of y=2x, except that the part which should bein quad III is actually in quad II.\nFor this graph, the vertex is when y=0. \nSo the vertex is at (0,0).
187414
2^y = 64/y\n\n\nthis is the intersection of two curves:\nt=2^y\nand\nt=64/y\n\nthat is an exponential and an equilateral hyperbola.\nPlotting the graphs you can have an idea of the intersection (in this case you have the exact solution by plotting integer values, but in general you may need to solve by iterative numeric formulas, after finding graphically a rough estimate of the solution)
27685
You have two equasions 2x=3y and y=2x+4\nYou're going to plug one equasion into the other; plug the y= into the  first equasion because the y is already by itself.\n2x=3(2x+4) (I plugged what y equaled into the y in the first equasion.\nThen you solve for x\n2x=3(2x+4)\n2x=6x+12\n-4x=12\nx=-3\nNow that you have solved for x, plug x back into the y='s equasion\ny=2x+4\ny=2(-3)+4\ny=-6+4\ny=-2
1169467
y = x^2 + 5x - 3\n2y=3x - 2\n----------------\n2y = 2x^2 + 10x - 6\n2y = 3x - 2\n-----------------------\n    0 = 2x^2 + 7x - 4\n=>2x^2 + 7x - 4 =0\n=>2x^2 + 8x - x - 4 = 0\n=>2x(x + 4) -1(x+4) = 0\n=> (2x - 1)(x + 4) = 0\ntherefore 2x - 1 = 0 => x = 1/2 or \nx + 4 = 0 or x = -4\nx = - 4 or 1/2  \ngot it?
985205
y = 2x^2 - 5    is a graph of parabola.\n\n-------------------------------------\n\nWhen x = # ,  this is an equation of a VERTICAL line.\n\nSimilarly,  when "x" is squared, "y" is to the power of one,\nthe parabola is a VERTICAL parabola. \n\nSo, now, we have established that the above equation is a VERTICAL parabola.\n\n-----------------------------------------\n\nA VERTICAL parabola is symmetric with respect to the y-axis, regardless of which direction the parabola opens (up or down).\n\n-----------------------------------------\n\nANSWER: This parabolic equation is symmetric with respect to the y-axis.
271990
x^2-2xy+4xz-8yz=\n(x^2-2xy) + (4xz-8yz)=\nx(x-2y) +4z(x-2y)=\n(x-2y)(x+4z)
844586
9x^2+y^2=9......(eqn no 1)\n3x-y=3 .............(eqn. no. 2)\nFrom equn. no. 2,we get y=3x-3=3(x-1)\nPutting the value of y in eqn.no.1,w get\n9x^2+{3(x-1)}^2=9\n=> 9x^2+9(x^2-2x+1)=9\n=>9x^2+9x^2-18x+9=9\n=>18x^2-18x=0\n=>18x(x-1)=0\nTherefore either 18x=0 or x-1=0\ntherefore x=0 or 1\nPutting the value of x in eqn no 2,we get 3X0-y=3 or y=-3  or 3X1-y=3,in which case y=0 \nTherefore x=0,y=-3 or x=1 and y=0
1151668
The x-coordinate decreased by 4 - (-2) = 6 and the y-coordinate decreased by 2 - 0 = 2\n-3 - 6 = -9\n5 - 2 = 3\nAnswer: Q'(-9,3)
979191
When converting from rectangular coordinates to polar coordinated, remember the following.\n\n@ = Theta\nx = r*cos(@)\ny =r*sin(@)\n\nr^2 = x^2 +y^2\n\ntan(@) = y/x\n\nso your equation\nx^2 +y^2 =36 becomes\nr^2 = 36   OR\nr =6 (which is a circle with radius 6)
453667
y=(x+8)^3-5\ny+5=(x+8)^3\n(y+5)^(1/3)=x+8\nx=(y+5)^(1/3)-8