Question

How do you find all points of inflection given `y=x^2/(2x+2)`?

Answer 1

For `y=x^2/(2x+2)`, there are no inflection points, but there is a vertical asymptote at `x=-1.`

Explanation:

Inflection points are the points where the graph's concavity changes—that is, where the slopes of the graph go from increasing to decreasing (or vice versa). Visually, inflection points are the points where the graph curves like a bowl on one side (concave up) and like a hill on the other (concave down). (e.g. `f(x)=sinx` has inflection points at every point where the graph crosses the `x`-axis—at `x=0,+-pi,+-2pi,`etc.)

Mathematically, inflection points occur where the second derivative is equal to 0. So we need to find `y''`.

For `y=x^2/(2x+2)`, we use the division rule to get

`y'=((2x)(2x+2)-x^2(2))/(2x+2)^2`

`color(white)(y')=(4x^2+4x-2x^2)/[2(x+1)]^2`

`color(white)(y')=(2(x^2+2x))/(4(x+1)^2)`

`color(white)(y')=(x^2+2x)/(2(x+1)^2)`

(We can leave the numerator un-factored to help us find its derivative again.)

Now, we take this `y'` and find its derivative: `y''`.

`y''=((2x+2)(2)(x+1)^2-(x^2+2x)(2)(2)(x+1)(1))/[2(x+1)^2]^2`

`color(white)(y'')=(cancel4(x+1)^(cancel(3)2)-cancel4x(x+2)cancel((x+1)))/(cancel4(x+1)^(cancel(4)3))`

`color(white)(y'')=((x+1)^2-x(x+2))/((x+1)^3)`

`color(white)(y'')=(cancel(x^2)+cancel(2x)+1-cancel(x^2)-cancel(2x))/((x+1)^3)`

`color(white)(y'')=1/(x+1)^3`

We now set this second derivative equal to 0 and try to solve for `x`...

`1/(x+1)^3=0`

Uh-oh! We can't do it. `1/"something"` can never equal 0, no matter what that something is.

Because of this, there are no inflection points, but there will be a vertical asymptote when `2x+2=0` (in other words, when `x=-1.)`