# `y = (x - 2)e^(-x)` Sketch the curve by locating max/mins, asymptotes, points of inflection, etc.

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### 1 Answer

`y=(x-2)e^(-x)`

(I) Asymptotes

To determine its horizontal asymptotes, take the limit of this function as x approaches positive and negative infinity.

`lim_(x->-oo) (x-2)e^(-x) =-oo`

`lim_(x->oo)(x-2)e^(-x)=0`

So, the function has only one horizontal asymptote. **Its horizontal asymptote is `y=0` **.

Moreover, there are no values of x which make the function undefined. * Thus, the function has no vertical asymptotes*.

(II) Maximum/Minimum Point

To determine the maximum or minimum point of a function, take its derivative.

`y=(x-2)e^(-x)`

`y'=(x-2)*e^(-x)*(-1) + e^(-x)*(1)`

`y'= e^(-x) ((-1)(x-2)+1)`

`y'=e^(-x)(-x + 3)`

Set the derivative equal to zero.

`0 = e^(-x) (-x+ 3)`

To solve for x, we only have to consider the factor (-x + 3). Since the first factor e^(-x) will never be zero for any value of x.

`0=-x + 3`

`x=3`

So, there is only one critical number, which is `x=3` .

To determine if the function is maximum or minimum at x=3, refer to the change of signs of first derivative before and after the critical number. If the change of signs of y' is from negative to positive, then, the function is minimum at that critical number. If the change of signs of y' is from positive to negative, then, the function is maximum at that point.

At the left of x=3, the interval is `(-oo, 3)` . And at the right of it, the interval is `(3,oo)` . Take a test value for each interval and plug-in them to the first derivative.

`y' = e^(-x)(-x + 3)`

For the interval `(-oo,3)` , let the test value be x=0.

`y'=e^(0)(0+3) = 3`

For the interval `(3,oo)` , let the test value be x=4.

`y'=e^(-4)(-4+3)=e^(-4)(-1) = -0.018`

Since the change of signs of y' is from positive to negative, then, the function is maximum at x=3. The value of y when x=3 is:

`y=(x-2)e^(-x)`

`y=(3-2)e^(-3)=0.05`

**Hence, the function has a maximum point (3, 0.05). And it has no minimum point.**

(III) Increasing/Decreasing Interval

To determine at which interval is the function increasing and decreasing, refer to the sign of the first derivative. If the sign of the first derivative is positive, the function is increasing at that interval. If the sign is negative, the function is decreasing.

At the left of the critical number, the interval is `(-oo,3)` . In this interval, the value of y' is positive.

And at the right of the critical number, the interval is `(3,oo)` . The value of y' here is negative.

**Thus, the function is increasing in the interval `(-oo,3)` and it is decreasing in the interval `(3,oo)` .**

(IV) Inflection point

To determine the inflection point, take the second derivative of the function.

`y'=e^(-x)(-x +3)`

`y'' = e^(-x)*(-1) + (-x+3)*e^(-x)*(-1)`

`y''=e^(-x)(-1 + (-x+3)(-1))`

`y''=e^(-x)(x-4)`

Set the second derivative equal to zero.

`0=e^(-x)(x-4)`

To solve for the value of x, consider only the factor (x-4). It is because the factor e^(-x) will never be zero for any value of x.

`0=x-4`

`4=x`

So the function changes concavity at x=4 only. The value of y when x=4 is:

y=(x-2)e^(-x)

`y=(4-2)e^(-4) = 2e^(-4)=0.04`

**Thus, the inflection point is (4, 0.04).**

(V) Concavity

To determine the concavity of the function, consider the region at the left and right of x=4.

At the left of x=4, the interval `(-oo,4)` . And at the right of it, the interval is `(4,oo)` .

Take a test value for each interval and plug-in them to the second derivative.

`y''=e^(-x)(x-4)`

If the second derivative is negative, the function is concave downward in that interval. If the value of the second derivative is positive, the function is concave upward.

For the first interval `(-oo,4)` , let the test value be x=0.

`y''=e^0(0-4)=-4 ` (Concave down)

For the second interval (4,oo), let the test value be x=5.

y''=e^(-5)(5-4)=0.007 (Concave up)

**Thus, the function is concave down in the interval `(-oo,4)` and it is concave up in the interval `(4,oo)` .**

(VI) Graph

**Therefore, the graph of the function `y = (x-2)e^(-x)` is:**