How do I find the slope for f(3) = 2?
115638
You can't.  f(3) = 2 doesn't fully define the function. Right now, you only have one point, (3,2). You need at least one other point, and then you can calculate the slope, assuming f(x) represents a line, by taking the difference in the values of f(x) and dividing it by the difference in the values of x.  Say for example you're also told f(5)=8.  Then the other point is (5,8), and the slope is (8-2)/(5-3)= 6/2 = 3.\n\nNow, if you mean the line f(x) = 2, the slope of that line is 0, i.e. a horizontal line, since the function is always 2 regardless of what x is.\n\nIf f(x) does not represent a line, iow you're doing a calculus problem, then you need to take the derivative of f(x), evaluate it at the point x=3, and that is the value of the slope of the tangent line at that point.
866549
The slope of a line is defined as:\n\nm = (change in y) / (change in x)\n\nor:\n\nm = (y2 - y1) / (x2 - x1)\n\nGiven the points (-3, 3) and (0,-3) means:\n\nx1 = -3\nx2 = 0\ny1 = 3\ny2 = -3\n\nPlugging these numbers into the equation for m yields:\n\nm = (-3 - 3) / (0 - -3)\n\nm = (-3 - 3) / (0 + 3)\n\nm = (-6) / (3)\n\nm = -2\n\nNote that a slope of 0 means that there is no change in y, or that y1 = y2
247082
Yes because to find slope you need change in y and change in x (or whatever coordinate system you use).\n\nSince you have both two x's and two y's then you're set.\n\nUse this formula: Δy/Δx where Δ stands for "change in" or (y1 - y2)/(x1 - x2)\n\nUsing the coordinates listed:\n(-2 - -2)/(6 - -8)=\n(-2 + 2)/(6 + 8)=\n(0)/(14)\n\nSince '0' divided by anything is zero, your slope is zero.\n\nThis would be a line that is parallel to the x axis 2 units below the axis.
32903
if you are given points (ex. (2,1) and (3,5) you can use this formula:\nm (slope) = y2-y1 / x2-x1\nso: 5 - 1 / 3-2 = 4/-1 or -4\n\nif you are given the equation of a line you can use this:\nx - ay = b\ny = -x + b / -a\nex.x-3y = 15\n-3y = -x + 15\ny = x/3 + 15\nso the slope is:\nx/3 = 1/3\nm(slope) = 1/3\nalways look for the coefficient of x and that is the slope.\n\nyou can also use this when you are given an angle:\nm (slope) = tan alpha
271844
If you need the tangent, take the derivative of the function at the specific point, which will give you the slope. Since you have taken the derivative at some specific point (a,f(a)) you also have a point. Now use the point-slope form to figure out the equation. Then simply get a graph paper and graph it.\n\nExample: f(x)=3tan(2x)\n\nf'(x)=6[sec(2x)]^2 (I assume you can take derivatives.)\n\nYou have a point (a,f(a))\n\nNow use the point-slope form: y-f(a)=f'(x)(x-a)\ny-3tan(2a)=6[sec(2a)]^2*(x-a) is the answer in this case. \n\nOr in function form: y=6[sec(2a)]^2*(x-a)+3tan(2a)\n\nTo graph it, find 2 points on the equation, then draw a straight line between them. That is your tangent.
963751
find the first derivative for the curve, and then substitute the x coordinate for the slope at that point.\n\nfor example, for the graph y = x^2, the first derivative is 2x. to find the slope at where x = 1, sub x = 1 into 2x and the slope at that point is 2.\n\nalternatively, draw a tangent to the point where u want to find the point slope and the slope of the tangent is the point slope. note that this method's accuracy depends on how accurately you draw your tangent
155791
Here is a good page describing slope.\n\nhttp://mathforum.org/cgraph/cslope/slope.html
911982
slope = f'(0),\nf'(x)= e^x 1/(x^2 +3) 2x + e^x ln( x^2 +e)\nf'(0)= e^0 1/3 (2(0)) + e^0 ln(e)\n= 1\n\nnow y=f(0)=e^0 ln( e) =1\nso the eq. of the tangent line is:\n\ny-1= x\ny=x+1\n\n\nfor question 2:\ne^{sin(0.001)} is very similar to e^0.001 which in turn is approx 1\ns
986272
f(x) = 3(2/x + 1)\ng(x) = x^2\n\n\nHope this helps.
907088
The slope is 1/3 and the point is (-2, -4).
450157
slope = (y2-y1)/(x2-x1)\n5= (k-4)/(2-3)\n5= (k-4)/(-1)\nk-4= (-1)*5\nk-4= -5\nk= 4-5\nso k= -1.\ni hope you understand they way I worked it out. If you have any more questions, pls feel free to ask me on: pinkieflava@yahoo.co.uk
846527
You'll have to do this in more than one step.  Basically it will sort of have to be a program.  I would start with the restart; command and then  define the function (I would look this up in the help menu since it is dependent on 2 variables).  Then use a command to find the derivative.  Then a command to solve for the pairs (x,y) when the derivative is 0.  That will give you 1.  For 2 and 3 you will also have to do a series of commands.  Use your calc book to refresh your memory on the procedure of finding these, it should be a similar procedure in maple.
820512
Y=MX+B\nso identify which number is next to the X, that will be your slope.  The other number is the Y-intercept.\n\nFirst problem:\nY=-2X+3\nthe -2 is next to the X so it is the slope (M)\nthe 3 is by itself so it is the Y-intercept (B)\nthe other two problems are the same way
604791
To find the slope, represented by m, you use the formula:  \n(the change in y)/(change in x)  \nor\n(y2-y1)/(x2-x1)\nSo for (2,-3) (4,5)  x1=2  x2=4  y1=-3  y2=5\n((5-(-3))/(4-2)=(5+3)/(4-2)=(8)/(2)=4\nso m=4\n\nTo find the distance, represented by d, use the formula:\nsquare root((x2-x1)^2 + (y2-y1)^2)\n\nfor (2,-3) (4,5) d= 8.25
407712
slope = -0.25\nintercept = 1
1066076
ok first set up the problem according to y=mx+b:\n3y+2x+6=0\n- you want to get y by itself so you subtract 2x+6 from both sides (or add -2x-6)\n- you should get 3y= -2x-6\n-then divide by 3(or multiply by 1/3)on both sides\n- you should get y=-2/3x-2 and since m = your slope then the slope is -2/3\n:D\nI LOVE MATH!!!!!!!!
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points