# y= (x^2)/ √(x - 4) Domain. restricting your attention only to values of x for which f(x) is defined. Intercepts. Find intersections of the graph with the axes. The Y-intercept is (0, f(0)). To...

- y= (x^2)/ √(x - 4)

**Domain.**restricting your attention only to values of x for which f(x) is defined.**Intercepts.**Find intersections of the graph with the axes. The Y-intercept is (0, f(0)). To obtain the X-intercept (c, 0) is harder and sometimes impossible. You need to find c such that f(c)= 0**Symmetry.**If f(x)=f(-x) , for example cos x = cos (-x), then the graph is symmetric with respect to the Y-axis. If f(x)= -f(-x), for example sin x= - sin (-x), then the graph is symmetric with respect to the origin.**Asymptotes.**Look for infinite limits and limits at infinity.**Intervals of increase or decrease and local maxima and minima**Reminder: Solve f’(x)> 0 and f’(x)<0 for monotonicity.

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### 1 Answer

1. You should notice that the radicand needs to be larger or equal to zero but since `sqrt(x-4)` is also denominator of the fraction `(x^2)/(sqrt(x-4)),` it cannot be equal to zero, such that:

`x - 4 gt 0 =gt x gt 4`

**Hence, the domain of the function is represented by the interval `(4,oo).` **

2. You need to find x and y intercepts such that:

`x = 0 =gt f(0) = (0^2)*sqrt(0-4) =gt f(0) = 0`

Hence, the graph of the function passes through origin (0,0).

`y = 0 =gt f(x) = 0`

You need to solve the equation `(x^2)/(sqrt(x-4)) = 0` .

Since `(sqrt(x-4))!=0` , then `x^2 = 0 =gt x = 0.`

Again, the fact that the graph passes through origin (0,0) is confirmed.

3. You need to test if the function is odd or even such that:

`f(-x) = (-x)^2sqrt(-x-4) =gt f(-x) = x^2sqrt(-x-4)`

**Hence, the function is not odd, nor even, hence, the graph of the function is not symmetric with respect to origin or y axis.**

4. You need to search for vertical asymptotes such that:

`lim_(x-gt4) (x^2)/(sqrt(x-4)) = 16/0 = +oo`

**Hence, the graph of the function has vertical asymptote when x tends to 4.**

You should search the horizontal asymptotes such that:

`lim_(x-gtoo) (x^2)/(sqrt(x-4)) = oo/oo`

You should use l'Hospital's theorem such that:

`lim_(x-gtoo) (4x)(sqrt(x-4)) = oo`

**Hence, the function has no horizontal asymptotes.**

You should search the oblique asymptotes `y = mx+n` such that:

`m = lim_(x-gtoo)(x^2)/(xsqrt(x-4))`

`m = lim_(x-gtoo) (x)/(sqrt(x-4)) = oo/oo`

m`= lim_(x-gtoo) 2(sqrt(x-4)) = oo`

**Since m=oo, hence, the function has no oblique asymptotes.**

5. You need to find derivative of the function using the quotient rule such that:

`f'(x) = (2xsqrt(x-4) - x^2/(2sqrt(x-4)))/(x-4)`

`f'(x) = (4x(x-4) - x^2)/(2(x-4)sqrt(x-4))`

`f'(x) = (4x^2 - 16x - x^2)/(2(x-4)sqrt(x-4))`

You need to solve the equation `f'(x) = 0` to find the minimum and maximum points of the function such that:

`f'(x) = 0 =gt 3x^2 - 16x = 0 `

`x(3x - 16) = 0 =gt x = 0`

`3x -16 = 0 =gt 3x = 16 =gt x = 16/3`

Notice that f'(x) is negative if `x in (0;16/3)` and it is positive if `x in (-oo;0)` or `x in (16/3 ; oo).`

**Hence, the function reaches its maximum at `x = 0` and it reaches its minimum at `x = 16/3` .**

**The function increases over `(-oo,0) ` and `(16/3,oo)` and it decreases over `(0,16/3).` **