Question

Find the x-intercepts and y-intercept of y=x^3-2x^2-4x+8

Answer 1

x intercept- (2,0) (-2,0)
y intercept- (0,8)

Answer 2

For the intersection with the x axis we have:x ^ 3-2x ^ 2-4x + 8 = 0Rewriting we have:
((x-2) ^ 2) (x + 2) = 0
The solutions are:
x1 = 2
x2 = -2
The intersections are:
(-2,0)
(2,0)
For the intersection with the axis and we have:
y = x ^ 3-2x ^ 2-4x + 8
y = (0) ^ 3-2 (0) ^ 2-4 (0) +8
y = 8
The point of intersection is:
(0, 8)
Answer:
The x-intercepts and y-intercept of y = x ^ 3-2x ^ 2-4x + 8 are:
(-2, 0) and (2, 0)
(0, 8)
respectively