Notice that ##x^2+4x = (x+2)^2 – 4## and take ##abs(x+2)## outside the square root to find two slant :

##y = x+2##

and

##y = -x-2##

Let ##f(x) = y = sqrt(x^2+4x) = sqrt(x(x+4))##

As a Real valued function, this has ##(-oo, -4] uu [0, oo)##, since ##x^2+4x >= 0## if and only if ##x in (-oo, -4] uu [0, oo).##

##sqrt(x^2+4x)##

##=sqrt(x^2+4x+4-4)##

##=sqrt((x+2)^2-4)##

##=sqrt((x+2)^2(1 – 4/((x+2)^2))##

##=abs(x+2) sqrt(1-4/(x+2)^2)##

As ##x->+-oo## we find that ##4/(x+2)^2 -> 0##, so ##f(x)## is asymptotic to ##abs(x+2)##

This results in two slant asymptotes:

##y = x+2## as ##x->+oo##

and

##y = -x-2## as ##x->-oo##

graph{(y-sqrt(x^2+4x))(y – x – 2)(y + x + 2) = 0 [-11.01, 8.99, -1.08, 8.92]}

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graph{y^2 + x^2 – 100 = 0 [-32.48, 32.44, -16.28, 16.2]}

This is a circle whose center point is the orgin and whose radius is 10 (##sqrt(100)##). The formula for a circle is ##(x-h)^2 + (y-k)^2 = r^2##.

In this case, h and k are 0 so the center is the orgin and the radius is ##sqrt(100)## = 10

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By using the center-radius form.The center is ##C## at ##(h,k)## meaning ##h = -4## and ##k = 2##.The equation will be ##(x – h)^2 + (y – k)^2 = r^2##That translates to:##(x + 4)^2 + (y – 2)^2 = 25##Expanding, we get##x^2 + 8x + 16 + y^2 – 4y + 4 = 25##or##x^2 + 8x + y^2 – 4y – 5 = 0##

For the graphing part, locate the center ##(-4,2)##. Open your compass 5 units to any direction, then make a circle. It should pass the following points (along with infinitely many other points):##(-4 + 5, 2)## ##->## ##(1,2)####(-4 – 5,2)## ##->## ##(-9, 2)####(-4, 2 + 5)## ##->## ##(-4, 7)####(-4, 2 – 5)## ##->## ##(-4, -3)##

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I think you could do it like this:

Let f(x) be the odd function we’re gonna work with. So, f(-x) = -f(x), By the definition of the derivative:

f'(x) = ##lim_(h->0)(f(x+h) – f(x))/h##.

We want to prove that f'(-x) = f'(x). Now, f'(-x) is:

##f'(-x) = lim_(h->0)(f(-x+h)-f(-x))/h## .

Since f(x) is and odd function, f(-x) = -f(x), therefore, we can rewrite the above expression as:

##f'(-x) = lim_(h->0)((-f(x-h)) + f(x))/h## = ##lim_(h->0)[-(f(x-h)-f(x))/h]##.

But we can rewrite the above expression as:

##f'(-x) = lim_(h->0)(f(x-h)-f(x))/-h##.

To make things clearer, let t = -h. Therefore:

##f'(-x) = lim_(t->0)(f(x+t)-f(x))/t##, which is just f'(x). That is: f'(-x) = f'(x). And there we have it: if f(x) is an odd function, then f'(x) is an even function.

Hope it helps you!

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How do you see the future of Advocacy on National level and explain role of the human services worker in these processes/field.

Future of Advocacy on National level• All the National governments will fulfil their responsibility for formulating andimplementing national policy honestly. Rights are equal for all. •…

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