Find the coordinates of the vertex for the function f(x)=-3x^2+6x-2?
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)
373083
Well, I'm not going to answer the question for you, but I'll tell you how to do it.\n\nWhen you're graphing the function, you set up a table with f(x) and x as the two columns, then solve for x for any set of numbers, like -3, -2, -1, 0, 1, 2, 3, and that's probably more than enough. Graph those on a piece of graph paper (f(x)=y, and then x). \n\nNow, to find the INVERSE, you simple switch the X and the Y. Instead of solving for x, you solve for f(x). So, your problem would be written x= -2/3y + 5. An easier way to do it, take the table you made before, and switch the columns, so that the solutions for x are underneath f(x) and the numbers you picked for f(x) are underneath the x column. Graph those coordinates. When you look at the graph, the inverse should be a mirror reflection of the first one, with  the graph of y=x as the line of symmetry. \n\nThat's how you find the inverse of any problem. Sorry if anything's unclear, if you need a better visual, I'm sure you could google up this exact method.
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)
491024
y = f(x) = 3/4x + 11/4
873165
f(g(x)) = (g(x)+4)^2 = (8x-8+4)^2 = (8x-4)^2 = 64x^2-64x+16\n\nThat's your composite function.\n\nHope this helps.
156411
If you are solving f(x) = ax^2 + bx + c = 0 by graphing, then there are three possibilities:\n\n1) graph of f(x) does not touch x-axis. Then there are no real solutions. Your real graph (x and y axis) does not tell you the solutions. You will have to do this algebraically (plug a, b, and c to quadratic formula). you will notice that you have to take the square root of a negative number. that is why these roots are not real numbers. they are both complex numbers.\n\n2) graph of f(x) touches x-axis at one point. Then there is one solution (with multiplicity of 2) at this x value. also, the quadratic polynomial factors into a perfect square.\n\n3) graph of f(x) touches x-axis at two distinct points. you can find the two solutions by their values on the x-axis.
449247
f(x)=(3x^2)-x+10\ng(x)=1-20x\n\n(f°g)=f[g(x)]\n\nSo we have:\n\nf[g(x)]={[3(1-20x)^2]-(1-20x)+10}\n={3[1-40x+(400x^2)]-1+20x+10}\n=3-120x+(1200x^2)-1+20x+10\n=(1200x^2)-100x+12\n\n(f°g)(5) means that we substitue 5 for x.\n\n[(1200x^2)-100x+12](5)\n={[1200(5)^2]-[100(5)]+12}\n=[1200(25)]-500+12\n=30000-488\n=29512
928287
y = int from 1 to 6 [f(x) - g(x)] dx\ny= int 1,6 [x^2-x-4] dx - int 1,6 [x^2-x-6] dx\n\nint 1,6 [x^2-x-4] dx\n[(x^3)/3] - [(x^2)/2] - 4x [1,6\ninputing 6 and 1 for x you get\n[216/3] -18-24 -([1/3] - [1/2] - 4) = 34 1/6 or 205/6\n\nint 1,6 [x^2-x-6] dx\n[(x^3)/3] - [(x^2)/2] - 6x\ninputing 6 and 1 for x you get\n[216/3] -18-36 -([1/3] - [1/2] - 6) = 24 1/6 or 145/6\n\nso you get 34 1/6 - 24 1/6 = 10
350103
Here's what you do:\n\nYour equation is y= 3/2x^2 - 6x + 5\n\nNotice that:\n\nWhen x = 1, y = 1/2\nWhen x = 2, y = 6 - 12 + 5 = -1\nWhen x = 3, y = 27/2 - 13 = 1/2\n\nSo, you can figure out that the axis of symmetry is x = 2 and you can see that the vertex is a minimum, right?
989452
The parent function is y = 6^x. That function contains the point (0,1). Since f(x) = 6^(x+1) is a translation of 6^x one unit to the left, the point (0,1) maps to (-1,1).  \n\nThe graph of f(x) = 6^(x+1) looks like the graph of 6^x translated one unit to the left.
986609
y=(x+3)^2=x^2+6x+9\ny'=2x+6=0\n2x=-6\nx=-3\ny(-3+3)^2=0\nso the vertex is (-3,0)
749496
By using the equation -b/2a you can get the x coordinate of the vertex and then plug that answer into the original equation to fond the y component.\n\ny=ax^2-bx+c, -b/2a\ny=2x^2-8x+15, -(-8)/(2*2)=2, plug 2 in for x and y=7. So vertex is at (2,7)
949784
y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18
692865
V = x * (4-x) * (6-x) = x^3 - 10x^2 + 24x\n\n0 < x < 4\n\nGraph the function and find the max between 0 and 4
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
735529
f(x)=-2x^4+6x+7\nf(-x)=-2(-x)^4+6(-x)+7\n=-2x^4-6x+7\nnether odd nor even