Respuesta :

Nirina7
the graph of y=-0.5 sqrt (x-3)+2
Df= {x/x-3>=0}
Df= [3, + infinity[
derivative of f(x)
f'(x)= -0.5 x 2 /sqrt (x-3)= - 1/sqrt (x-3) <0, f is a decreasing function for all x in the Df
limf(x)=2 x--------->3,  limf(x)=-infinity, x--------->+infinity

look at the graph
Ver imagen Nirina7

Answer:

The function is given to be :

[tex]y=-0.5\sqrt{x-3}+2[/tex]

Since, square root function cannot have negative value for any real value of square root

⇒ x - 3 > 0

⇒ x > 3

At x = 3 , y = 2

So, graph of the given function will have least value of x = 3 and highest value of x = ∞

Least value of y = -∞ and highest value of y = 2

Also, at y = 0, x-intercept , x = 19

Hence, the required graph of the given function is given below :

Ver imagen throwdolbeau

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