Question #fbdc5

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Answer 1 · 2017-04-25T16:05:11

#x>=0#
graph{(sqrt(45x^2)) [-6.67, 5.814, -1.274, 4.966]}

Assume that variables represent nonnegative real numbers.
So, #x# can't be negative.

#sqrt(45x^2)#

All we need to do is give #x# any positive value we want.

We can do #0#.

#sqrt(45x^2)#

#sqrt(45(0)^2)#

#sqrt(45(0))#

#sqrt0#

#0#

This means our point on #x# is #0#
and our point on #y# is #0#
#(0,0)#

Let's try #2# as well.

#sqrt(45x^2)#

#sqrt(45(2)^2)#

#sqrt(45(4))#

#sqrt(180)#

#~~13.416#

This means our point on #x# is #2#
and our point on #y# is #13.416#
#(2,13.416)#

Keep doing this until you have all of the positive #x# values that you need to draw your graph.

Do not draw the part of the graph with negative #x# values because that is the rule from our question.

Your graph should look like the answer I gave you, but without the line on the left side of the y-axis. Thos are negative #x#.

#x>=0#