the graph of y=3x^2 would be shifted up, down, left, right; thinner or wider from/than the graph of y=x^2?
923158
Thinner simple work its cause the 3 is greater than the one other than drwing the grph there is really no work. And if u need the graph then just make an X Y table using values -3 to 3 then solve for y and gragh them
670280
If you have a parabola in the form of y = a(x - h)^2 + k, then you can read off a lot of your particulars directly from the equation:\n\nvertex: (h, k)\ndirection: for a>0, parabola opens up, for a<0, parabola opens down\naxis of symmetry: x = h\n\nSo you have a parabola that opens down with vertex at (-2, 3), and axis of symmetry at x = -2.\n\nNow compare this to y = x^2: opens up, vertex is at (0, 0), and axis of symmetry at y = 0.  Your first equation will also be skinnier because of the a value: if |a|<1, your parabola is wider than y = x^2, and if |a|>1, then your parabola is skinnier than y = x^2.
405955
If you are reflecting across the y axis, x becomes -x.\n\nIf you are reflecting across the x axis, y becomes -y.\n\nIf you are reflecting about the origin, both x and y change sign.\n\nIs that what you are looking for?
844451
to the right, i.e. x-> infinity,  3x^2 - 5x -7^-x is more like 3x^2,\nnow , to the left, ie x-> -infinity, 3x^2-5x-7^{-x} is more like the exponential -7^{-x} \nis this what you mean, may be if you provided an example, i could help you better.\n'
817379
y=(x^2-3x)^2\ny=x^4-9x^2\nAnd then I'm confused.
113113
Think of this as an even power function.  Look at the graph of y=-x^2.  Looks like an arch.  So does y=-x^4.  Now you just need to calculate the other parts of this function to match your initial, final, and maximum.  The expression you will be using is y=-(ax+c)^2+b.  The 'a' will change the "roundness" and the 'b' will change the maximum.  The 'c' will change the horizontal position.
824521
this just means Y is a function of x\n\ny = x^2(2-3)\ny = -x^2
837061
it can be the original way you had it (c=mx +y) but it is alot simpler in the long run to rearrange it to (y=mx+c) (y being y, x being x, m being gradient, c being a constant). this way it is easyer to solve.
788027
The graph is a V.  From the 0,0 draw a line with slope = 1 to the right of the y axis and draw a line with slope = -1 to the right of the y axis.  For y<|x|.  All you have to do is shade in the area below the V.  If y>|X|, all you have to do is shade in the area above the V.  Less than, shade below; greater than, shade above.  That should answer all the questions you've submitted.\n\nAnd no, absolute values do not form parabolas.  Some do, not all.  Like y = sqrt[|x-1|].  This forms 2 graphs each going away from each other.\nNice going NerdyKid.  You're not a nerd, you're a dork.
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.
57318
9:3 simplifies to 3:1\n\ndivide both by 3 to get the answer
406488
If the number in front of x^2 is positive it opens up.\nIf it is negative it opens down.
735129
they would look like cross (X).. the sin(-x) graph loos like  / and     -sinx would look like &#92;. \n\nIf you dont get it please visit this figure uploaded on my webpage\n\nhttp://www2.lut.fi/~vasava/untitled.jpg.
942299
i cannot do this for you, but if it is e^(x^2) then it is going to be symmetric along the y axis right?\n\nso take the graph of e^u, where u=x^2, and do it from 0 to 9 (the square of 3) and then reflect that on the y axis and you have your answer. At u=0 the graph is going to equl to (0,1) and at 9 it will look like (9, e^9).  If you wanted to go from e^(x^2) from our e^u then it will just be squished (closer to the y axis at all points except 0), but you will still get a general feeling from what i said.\n\nalso i think it is possible you left out a negative in front of the x in which case the graph will look like the normal curve: http://www.uow.edu.au/student/attributes/statlit/modules/images/normal-curve-1.gif
787754
y = -x^2 will have a vertex at (0,0).  Which way will the parabola move if you add 4 to the x?  Just move the vertex four units that way.
748852
yes the signs have just been changed\ntake -x+y=-2 and multiply by -1 on both sides\n \nyou get x-y=2 and since the same thing was done to both sides the equation is equivalent
152511
Which function, Subsitution, Linear COmbination, Graph?