Graph:  y=e^(x^2) ; -3≤x≤3?
913432
It is symmetric around the y axis. It increases exponentially as x gets further from zero and follows this trend forever. At x=0, y=1 because anything (except 0) raised to the 0 power equals 1. This thing gets really big really fast.
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
152046
y=|x+2| \n this shifts the "v" shaped graph 2 units to the left from the origin.
338193
The equation represents a straight line graph whose gradient is 3 , and intercepts the the ordinate axis(i.e y-axis) at the -2 position.\n\nthe easiest way to arrive at this is to draw a table of values. Such that you find the values of y, using values of x from -6 to +6.\n\nYour graph will confirm my story, \nBut from mere looking at the equation I can make a good prognosis, when i compare it with Y=mx + c .where Y represents the value of the equation, X represents the value of x which made y what it is.\nM stands for the gradient of the graph, while c stands for the intercept on the Y-axis.
793059
give values of -3,-2,-1,0,1,2,3 to x and find the corresponding values of y\nyou get the ordered pairs\nnow plot
817379
y=(x^2-3x)^2\ny=x^4-9x^2\nAnd then I'm confused.
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
861221
looks like the vertex is (-5, 3)
223761
if your equation is y=x^2, then the graph that you draw is going to be a parabola. You can see this by having a table first. The table should consist of 2 values...x and y. So you say when \nx=-3, y=9,\nx=-2, y=4,\nx=-1, y=1\nx=0,  y=0\nx=1,  y=1\nx=2,  y=4\nx=3,  y=9\nso now if you have a square line paper, just plot it and you will see that it will give you a parabola. Another thing is that when you draw it, first just plot the points and then when you connect the point connect them using curves and not a line,...that's why u get a u shaped graph and not just a combo of lines.
621381
b) Replace y with 0\n(100+x)^0 = 2*100^0\n1 = 2*1\n1 = 2\nThis implies that y cannot be zero.\nc) Replace x with 0.\n(100+0)^y = 2*100^y\n100^y = 2*100^y\n0 = 2*100^y - 100^y\n0 = 100^y\nThis is never true.\nThis implies that x cannot be zero.\na) The graph looks similar to the graph of y = 1/x
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
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).
411757
since y=+-(x)^(3/2) Range [0,oo[\nnP=(n,n*n^1/2) and x_n/y_n= [(n^1/2^)/n]=n^-1/2
967981
y'=cos(x^2e^x)(2xe^x+x^2e^x)\n\ny''=-sin(x^2e^x)(2xe^x+x^2e^x)^2+\n      (2(e^x+xe^x)+(2xe^x+x^2e^x))cos(x^2e^x)
241669
Y=x^3(x+2)(3x-1)\n we have three possibility for x\nx^3=0 than X=0\nx+2=0 than X=-2\n3x-1=0 than x=1/3\n\nso the answer is a
931775
3e^x. Pull the constant outside the integral sign\nand just let it sit there!
837305
f(x) = [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2) ]\n= 3(x - 2)/(x - 3) (x ≠2, -2)\n= [3(x - 3) + 3]/(x - 3)\n= 3 + 3/(x - 3)\n\n1. Lim x→2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→2 [3 + 3/(x - 3)]  \n= 0 \nand \nLim x→-2 [ 3(x+2)(x-2)^2 ] / [ (x-3)(x+2)(x-2)]\n= lim x→-2 [3 + 3/(x - 3)] \n= 13/5\nSo it has 'holes' at (2, 0) and (-2, 13/5)\n\n2. It is undefined when x = 3, 2 and -2\nSo domain is all x except x = 3, 2, -2\nRange all y except y = 0 and 13/5\n\n3. As x increases and decreases without bound \n3/(x-3) approaches 0\nso 3 + 3/(x-3) approaches 3\n\nie lim x → ±∞ [3 + 3/(x-3)] = 3\n\n4. When x < 3\nx - 3 < 0 \nSo 3 + 3/(x-3) < 3\n\nWhen x > 3\nx - 3 > 0 \nSo 3 + 3/(x-3) > 3\n\nSo for x > 3 the curve is above y = 3 and for x< 3 the curve is below y = 3\n\n5. When y = f(x) = 0\n3(x - 2)/(x - 3) = 0\nie x = 2\nSo f(x) cuts x-axis at x = 2\n\n6. When x = 0\nf(x) = 3*(-2)/(-3) = 2\nSo y-intercept is y = 2\n\n7. It is rotationally symmetrical about the point (3, 3) (order 2) except for the 'holes'.\n\nThat should get you a good start I think