Question

How do you find intercepts, extrema, points of inflections, asymptotes and graph `y=2-x-x^3`?

Answer 1

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Explanation:

To find intercepts:

x intercept means "the point where the graph cuts x-axis".At this point the y co-ordinate will be 0(the point will have form `P(x_0,0)`, where `x_0` is its point of x intercept)
similarly, at y intercept x=0;
now put x=0 in equation to find y intercept
and put y=0 in equation to find x intercept

from the question,
`y=2-x-x^3`
`=>` at `x=0, y=2-(0)-(0)^3`
`=>y=2` `...` `(` at `x=0)`
thus y intercept is at point `(2,0)`
similarly putting y=0 we'll get x intercept(s)
`0=2-x-x^3`
`=>(x-1)(x^2+x+2)=0`
`x-1=0` or `x^2+x+2=0`
`=>x=1,` and no real solution of the equation `x^2+x+2=0` exist
`:.` x intercept is at point `(1,0)`

for extrema and points of inflexion use derivative method
the point(s) where the first derivative of the equation is 0(or the points where the slope of curve is 0) are called critical point(s)
on differentiating `y=2-x-x^3`, we get
`dy/dx=d/dx(2-x-x^3)`
`=>dy/dx=-1-3x^2`, putting `dy/dx=0`, we'll get critical points as
`-1-3x^2=0`
`=>3x^2=-1=>x^2=-1/3...i.e` not possible(as square of a number cannot be negative),
hence the curve will have no critical point
also as `=>dy/dx=-1-3x^2<0`
the function`y=2-x-x^3` is ever decreasing and have maximum at `x=oo` and minimum at `x=-oo`.

for points of inflexion (the point where curve changes concavity or shape) the second derivative must be 0
we have second derivative as
`(d^2y)/dx^2=d/dx(-1-3x^2)=-6x`
`(d^2y)/dx^2=0 =>x=0` ,at x=0, y=2
`:. (0,2)` is point of inflexion

since, `` is not of the form `(p(x))/(q(x))`,(called the rational form)
it has no asymptotes
it will be more clear from the graph