find zeros of a function calculator
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Answers:hit the top left button: y=, then type 7x + 4 into it, then hit the top right button, graph, then hit trace, and move to the left and right until x is equal to zero (you'll know because it shows the coordinates on the bottom)
Answers:try writing it as a factor:) (x+a)(x^2 + bx + c) = 0 x^3 + ax^2 +bx^2 + abx + cx + ca = 0 x^3 + (a+b)x^2 + (ab+c)x + ca = 0 here: x^3+6^2+12x+8 ca=8 ab+c = 12 a+b = 6 c= 4,a=2, b =4 x^3+6^2+12x+8 (x+2)(x^2 + 4x + 4) = (x+2)(x+2)(x+2) so with x^3+6^2+12x+8 = 0 (x+2)(x+2)(x+2) = 0 x= -2
Answers:Get out a graphing calculator. Then hit "Y=". Then put in the equation above. Then, hit GRAPH. Then, hit ZOOM, then select option two, hit enter, then hit enter again to zoom in. Then, find where the line hits the x axis between one and two. Then, hit 2nd TRACE (CALC), and hit option two, ZERO. Then, using the arrow keys, select a place on the line to the left of the zero you are looking at, and hit ENTER. Then select a place to the right of the place where the line hits zero and hit ENTER again. Then, select about where the line hits zero, then hit ENTER again. At the bottom of your graph, you should see something like X=.22187616, and Y=0. Therefore, your answer, rounded to the nearest tenth, would be 0.2. Hope that helped.
Answers:If you know how to use synthetic division, then you're set. synthetically divide x = 1 in, to get (x - 1) (some degree 3 equation) and then synthetically divide x = -1, to get (x - 1) (x + 1) (some quadratic equation) and then you can factor the quadratic equation. If you don't know how to synthetically divide, check out: http://www.purplemath.com/modules/synthdiv.htm :)