find the slope of the secant line through the points (4, f(4)) and (4+h, f(4+h)) on the graph y=f(x)?
914503
You should remember how to do this in the future because the limit of this expression as h approaches 0 gives you the derivative or slope of f(x) at x = 4.\n\n[ f(4 + h) - f(4) ] / [ (4 + h) - 4 ]\n\n[ (4 + h)^2 - 4^2 ] / h\n\n(h^2 + 8h + 16 - 16) / h\n\nh + 8
1154399
First you need to find the slope of the line, by finding two points on the line and making a fraction found by subtracting the two y-coordinates and putting this over what you get by subtracting the two x-coordinates.  Then if you can, find the y-intercept.  The equation is y = mx + b where m = slope and b = y-intercept.\n\nSuppose your trend line contains the point (4, 20) and hits the y-axis at (0,8)\n\nThe slope is (20-8)/(4-0) = 12/4 = 3\n\nThe y-intercept is 8\n\nSo the equation is y = 3x + 8
877049
use the slope formula.\n\ny sub 2 minus y sub 1 divided by x sub 2 minus y sub 1\n\nthen you get 1-5 divided by 0-2\n\nthen you get -4 divided by -2\n\nthe answer is 2
1121252
slope = m\nperpendicular slope = (-1)/m\n\nslope = (y2-y1)/(x2-x1)\nslope = (2-5)/(4-2)\nslope = -3/2\n\nperpendicular slope = 2/3
994382
The quick answer is to take the derivative of f(x) ...\n\nf'(x) = 2 (2x - 4) 2 = 4 (2x - 4)\n\n... and evaluate it at x = 4\nto get a slope for the tangent line of\nf'(4) = 16.  [Ask if you need to know\nwhy / how to do this.]\n\nNow you want the equation of a line through (4, 16) with slope 16.  That would be\n\n(y - 16) = 16 (x - 4)\n\nor\n\ny = 16 x - 48\n\nCheck: it has slope 16 and passes through (4, 16).
746991
the slope is 4\nwhen an equation is in standard form:\ny=mx+b\nm (which is the coefficient of x) is the slope
35268
slope = (-4-(-6))/(-2-0)\nslope = -1
293210
the graph of |x - 2| + |y - 5| = 5 consists of four lines which are graphs of the following functions:\nx-2+y-5=5 which is  line A: x+y=12\n-x+2-y+5=5 which is line B: x+y=2\n-x+2+y-5=5 which is line C: x-y=-8\nx-2-y+5=5 which is line D: x-y=2\nnext thing to do is graph these four lines. the easiest way is to find the 0s. \nline A: x+y=12\n        x=0 y=12 point A1(0,12)\n        y=12 x=0 point A2(12,0)\nline B:  x+y=2\n         x=0 y=2 point B1(0,2)\n         x=2 y=0  point B2(2,0)\nline C: x-y=-8\n         x=0 y=8 point C1(0,8)\n         x=-8 y=0 point C2(-8,0)\nline D: x-y=2\n        x=0 y=-2  point D1(0,-2)\n        x=2 y=0   point D2(2,0)\nthese four lines enclose a figure that is a rectangle. lets find the length of each side. after you plot the four lines you will notice that line x+y=12 intersects with line x-y=2 lets call the intersection point I1 to find I1 coordinates we rearrange y=12-x and y=x-2 and from y=y we get x-2=12-x or 2x=14 or x=7 if x=7 y=12-7=7-2=5 the point I1 is I1(7,5).\nline x+y=12 also intersects with line x-y=-8. Lets call this intersection point I2. to find coordinates we have y=12-x and y=x+8 y=y or 12-x=x+8 or 2x=4 or x=2 for x=2 y=12-2=2+8=10 so the point is I2(2,10) \nside I1I2 of the rectangle length is side a. to find length of a we do\na^2=(7-2)^2+(10-5)^2=25+25=50\na=sqrt(50)\nwe need the second side, lets call it b.\nto find length of side b we need at least one more point. lets find intersection of x-y=2 and x+y=2 and call it point I3. y=x-2 and y=2-x or x-2=2-x or 2x=4 or x=2. for x=2 y=2-2=2-2=0 point I3(2,0)\nside b:\nb^2=(7-2)^2+(5-0)^2=50\nb=sqrt(50)\nit means a=b or our rectangle is a square!\nso the area of the square is Area=side^2 since our square&#92;s side is a=b=sqrt(50), the area is 50.\nI hope this helps, I really don't have time to explain it in more details.
1115688
To graph functions, first get a piece of graph paper or you can free hand the X-Y axis and plot the points between -5 and 5 on both axis.\n\n1) h(x) = 2/x  domain of x (-1, 0, 1, 2)\n\nFind the coordinate pairs for each member of the domain.\nh(-1) = 2/-1  -->  (-1, -2)\nh(0) = 2/0  -->  undefined so no point on the graph\nh(1) = 2/1  -->  (1,2)\nh(2) = 2/2  -->  (2,1)\n\nThe graph will be the three points above.  It is not a line since the domain is limited to the integers listed.\n\n\n2) f(x) = x-1  domain is x/x is a real numbers (which excludes x=0 from the domain, but includes all other real numbers)\n\nSince the equation is in slope-intercept form, i.e.  y = mx + b\nthe slope of the line is m=1 and the intercept on the y-axis is b=-1\n\nYou can draw the line that way, or determine two points on the line like this:\nf(-1) = -1 - 1  -->  (-1,-2)\nf(1) = 1 - 1  -->  (1, 0)\n\nSince the domain is the real numbers except 0, a solid line is drawn to extend beyond both points.  Either way you draw the line, make an open point (ie. a circle) on the line where x=0.  Don't draw the line through the open point.  The line is terminated with arrows to indicate that it extend to infinity in both directions
1106036
Rise over run, or (y2-y1)/(x2-x1).  If your first point was (x1,y1) and your second point (x2,y2), then it's easy:\n\n[8 - (-4)] / [14 - 10] = (8+4)/(14-10) = 12/4 = 3\n\nThe slope of the line is 3.
846578
1.  Since f(x) = x² + bx + c, and you know f(4) = 9 (since (4,9) is a point on the parabola, specifically the vertex), you can solve for f(4) = \n\n16 + 4b + c = 9, so 4b + c = -7\n\nSince (4,9) is the vertex, you know the tangent at this point is zero, i.e., f '(4) = 0.  (I hope this is a calculus HW problem so that you know what I am talking about!)\n\nSo get f '(x) = 2x + b\n\nSince f '(4) = 0, we have 2(4) + b = 0, so b = -8\n\nSince b = -8, and 4b + c = -7, we can now solve for c:\n\n4(-8) + c = -7, so c = 25\n\n----------\n\n2.  Not sure I can help here.  I took the first and second derivatives of h(x) to determine that the function has a local min at x=7/3 and a local max at x=-1, but h(7/3) ≠ -5 and h(-1) ≠ 2, so not sure where to go from here, sorry!\n\n----------\n\n3.  A parabola has a 2nd-degree polynomial equation, or y = ax² + bx + c.  We know two values, namely (4,4) and (2,8) so we can set up two equations like thus:\n\n4 = a(4²) + b(4) + c\n8 = a(2²) + b(2) + c\n\nwhich we can rewrite as:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n\nRight now we have 3 unknowns and only 2 equations, but we can get a third equation by taking the first derivative, or f '(x) = 2ax + b.  Since the vertex is at (4,4) we know that f '(4) = 0, and we can get a third equation:\n\n0 = 2a(4) + b = 8a + b\n\nSo now we have 3 equations in 3 unknowns and we can solve for a, b, and c:\n\n4 = 16a + 4b + c\n8 = 4a + 2b +c\n0 = 8a + b\n\nSubtract equation 1 from equation 2, then divide by 2 to get:  \n2 = -6a - b\n\nNow add equation 3 to this to get 2 = 2a, or a=1.\n\nNow let's back-substitute to get b and c:\n\n2 = -6(1) - b, so b = -8\n\n8 = 4a + 2b +c\n8 = 4(1) + 2(-8) + c, so c = 20\n\nSo the equation of our parabola is:\nf(x) = x² -8x + 20\n\n(Double check to make sure f(4)=4 and f(2)=8).
910261
Slope = m\n= y2-y1&#92;x2-x1 \n= -4-6/-3-2\n = -10&#92;-5\n=2\n\nnote: / means divided by\n\nthe slope of y= 2&#92;3x-4 is 2&#92;3
985362
x-4y+4=12\n\nyou want to get it in y= mx + b form \n\nso switch the 4y and the 12 \n\nx + 4 -12 = 4y\nx - 8 = 4y\n\nnow divide by 4\n\n(1/4)x -2 = y\n\nso the slope is 1/4 and it crosses the y axis at -2\n\nsome points are \n(0, -2)\n(-4, -3)\n(-8, -4)\n(4, -1)\n(8, 0)\n(12, 1)\n\nhope this helped\nbest of luck
805503
need to take the derivative vector:\n( -3 sin t, 4 cost)\nwhen t= pi/4, the vector is: \n(- 3 sin(pi/4), 4 cos(pi/4)) = ( -3/sqrt(2), 4/sqrt(2) )\nnow, the slope of the tangent line is:\n (4/sqrt(2) ) /  ( -3/sqrt(2) ) =-4/3\nso the equation of the tangent line is:\ny- 4sin(pi/4) = -4/3 (x-3cos(pi/4)) \ny-4/sqrt(2) = -4/3( x-3/sqrt(2) )\n'
856807
Yes that will be correct because after you isolate the (y) in other words put it in slope inercect form it would be \ny=2x+4 \nm=2\nb=4
771497
No - the slope is going to be a number, not an equation.\n\nd/dx (2x^2) = 4x\n\nAt x = -2, the slope will be -8.\n\nThe equation of the tangent LINE would be\n\ny + 8 = -8(x + 2)\n\ny = -8x - 24, as you have it.\n\nAnd no - getting the terminology right is not a minor point.
679595
I can't really graph anything here, but to do so, find the intercepts.\n\nWhen x = 0 (the line crosses the vertical axis) then y = 4\nWhen y = 0 (the line crosses the horizontal axis) then x = 1\n\nSo, the line will cross the vertical (y) axis at 4 and the horizontal (x) axis at 1.\n\nNow for the shading. Consider that the point 0,0 is below the line you just drew. Plug that into your equation. 4*0 + 0 > 4 ... or 0>4. Clearly this is false. So 0,0 isn't part of your solution. Therefore, the shading must be above the line.