How do you graph #y=-sqrt(4-x^2)#?

1 Answer
Feb 29, 2016

Represents a semi circle, whose circumference lies below #x#-axis and which has origin #(0,0)# as its centre.

Explanation:

Plotted graph is as below.
graph{y=-sqrt(4-x^2) [-5, 5, -2.5, 2.5]}

Given expression is
#y=-sqrt(4-x^2)# ..........(1)
If we square both sides we obtain

#y^2=4-x^2#, rearranging we obtain

#x^2+y^2=2^2#.......(2)
It looks like an equation of a circle.
General equation of a circle whose center is at the point #(h,k)# and radius #r# is
#(x-h)^2+(y-k)^2=r^2#

So the equation (2) is of a circle which has radius #r=2#, and origin #(0,0)# as its center.
From the given expression we deduce that

  1. Equation (1) is a curve which has a properties as above. Also it must satisfy following two conditions.

  2. That #y# always has negative values due to the presence of #-ve# sign on the right hand side term.

  3. As square root of any negative number is imaginary and therefore, can not be plotted on a #x,y# graph. Implies that, argument of square root term must be positive.

Mathematically it can be written as
#4-x^2>=0#

Taking #x# to the left hand side and taking square root of both sides
we obtain #x<=2#

We see that the equation (1) represents a semi circle, whose circumference lies below #x#-axis and which has origin #(0,0)# as its center and radius #r=2#