Graph of f(x) = (x^3 - 5x ^2 + 6x)/(x^2 - x- 2). Can anyone please post a picture of this or something?
2 answers
go to wolframalpha dot com and put in (x^3 - 5x ^2 + 6x)/(x^2 - x- 2), it will show you a plot
We don't draw graphs here. I suggest you simplify it first by factoring.
It should be hard for you to graph this yourself. Just plot a dozen points or so that you can calculate easily.
(x^3 - 5x ^2 + 6x)/(x^2 - x- 2)
= x(x-6)(x+1))/[(x-2)(x+1)]
= x*(x-6)/(x-2)
When x = -20, y = -23.6
When x = -15, y = -18.5
when x = -10, y = -13.3
When x = -5, y = -7.85
When x = 0, y = 0
When x = 2 (left side) y = +infinity
When x = 2 (right side)y = -infinity
When x = 6, y = 0
When x = 10, y = 5
When x = 20, y = 15.5
The graph will have a vertical asymptote at x=2, and for very large positive and negative x, it will be asymptotic to the line y = x.
It should be hard for you to graph this yourself. Just plot a dozen points or so that you can calculate easily.
(x^3 - 5x ^2 + 6x)/(x^2 - x- 2)
= x(x-6)(x+1))/[(x-2)(x+1)]
= x*(x-6)/(x-2)
When x = -20, y = -23.6
When x = -15, y = -18.5
when x = -10, y = -13.3
When x = -5, y = -7.85
When x = 0, y = 0
When x = 2 (left side) y = +infinity
When x = 2 (right side)y = -infinity
When x = 6, y = 0
When x = 10, y = 5
When x = 20, y = 15.5
The graph will have a vertical asymptote at x=2, and for very large positive and negative x, it will be asymptotic to the line y = x.
1 answer