Area of Rectangle As Function of x

See attachment.

I say the area in terms of x and y is
A = length • width or x•y.

I know that y = sqrt{36 - x^2}.

A(x) = x•y

Replace y with sqrt{36 - x^2}.

A(x) = x•sqrt{36 - x^2}

Is this function?

If not, why not?

See attachment for domain.

I say the domain is (-00, -6] U [6, 00).

Yes?

 

A(x) = x•sqrt{36 - x^2}=> correct


=> x=6 or x=-6=> points are (6,0) and (-6,0)-> so, both are in domain
so, the domain is is [-6,6]
or
{x element R : -6<=x<=6}

 

The domain is on the line y = 0 from -6 to 6.

 

The domain is on the line x from -6 to 6 (take a look at graph)

 

The x-axis is the line y = 0.

Yes?

 

correction to the formula for area, there was a mistake

length= x units left fro origin and x units right from origin: so L=x+x=2x

width=W=y=sqrt(36-x)^2

we know that area of rectangle is length*width

A=L*W

now, we can plug values

A=2x*sqrt(36-x)^2 see attached to understand why is 2x in formula

 

My equation is missing number 2. Your graph makes clear why 2x and not x alone is multiplied by
sqrt{36 - x^2} to form the needed formula.

P. S. If you want to explore the other site on your own (without me), feel free to do so. Members jonah and romsek have no beef with you. As you already know, I will stay here for the rest of precalculus. Look for precalculus questions here. I see no reason why you should limit your contacts with math friends just for me. Be back tomorrow with more precalculus questions.

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Note 2:

I also have decided to join calculusworkshop (yearly plan) to watch precalculus videos by math expert Jenn. I am going to do everything within my power to "master" or better yet, increase my knowledge of this wonderful course.