Same thing as they do when you are graphing.
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solutions are the well solution to the problem. X-intercepts are wherever a graph cross the x axis, which are hte solutions when you have to find out what x is, zeros are pretty much the same thing although i think that include y-intercepts as well..... not sure. and roots are the same thing as x-intercepts. so they are all more or less the same thing
In general, there is no relationship.
(1,-36); -5 and 7y = x2-2x-35; f(x) = ax2+bx+cvertex, method 1:There is a formula for the vertex, if you can remember it. (h,k) = [-b/2a, f(-b/2a)].-b/2a = 2/2x1 = 2/2 = 1. f(1) = 12-2x1-35 =1-2-35 = -36. The vertex is (1,-36).vertex, method 2:Otherwise you complete the square to convert to the vertex form, y = a(x-h)2 + k. Complete the square for x2-2x: the constant term will be (-2/2)2 = (-1)2 = 1.y = (x2-2x+1)-1-35 = (x-1)2 -36 ==> The vertex is (h,k) = (1,-36).x-intercepts, method 1:To find the x-intercepts, let y=0: 0 = x2-2x-35 (x-7)(x+5) = 0 x-7=0 or x+5=0 x=7 or x=-5. The intercepts are -5 and 7.x-intercepts, method 2:There is also a nice trick to find the intercepts: 36 = 62. the intercepts are 1+/- 6 = 7 and -5.
x: x = 15 y: y = 10 z: z = -5
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