function graphs

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August 6th 2009, 06:47 PM
live_laugh_luv27

function graphs

How would I graph these two functions on a calculator?
I need to find domain, range, zeros, etc., along with a sketch of graph. Or would it be easier to graph by hand? how? Thanks!

$f(x) = log_2x$

$f(x) =\sqrt{a^2-x^2}$
August 6th 2009, 07:01 PM
artvandalay11

Quote:

Originally Posted by live_laugh_luv27

How would I graph these two graphs on a calculator?
I need to find domain, range, zeros, etc., along with a sketch of graph. Or would it be easier to graph by hand? how? Thanks!

$f(x) = log_2x$

$f(x) =\sqrt{a^2-x^2}$

The first one can be put in as $\frac{\log{x}}{\log{2}}$ into the calculator

The second one is no good with a calculator because there are two variables. Sketch it by hand and you can find intercepts that depend on the value of a (y=0 at x=a,-a and x=0 when y=a). You can find the domain by making sure the what is under the radical is $\geq 0$ and you can find the range by solving the equation for x, and then finding the domain of that new function (you are finding the domain of the inverse function)
August 7th 2009, 11:01 AM
dhiab

Quote:

Originally Posted by artvandalay11

The first one can be put in as $\frac{\log{x}}{\log{2}}$ into the calculator

The second one is no good with a calculator because there are two variables. Sketch it by hand and you can find intercepts that depend on the value of a (y=0 at x=a,-a and x=0 when y=a). You can find the domain by making sure the what is under the radical is $\geq 0$ and you can find the range by solving the equation for x, and then finding the domain of that new function (you are finding the domain of the inverse function)

Hello : thank you
I"think a is the parameter no variable.(Evilgrin)
August 7th 2009, 11:33 AM
Amer

Quote:

Originally Posted by live_laugh_luv27

How would I graph these two functions on a calculator?
I need to find domain, range, zeros, etc., along with a sketch of graph. Or would it be easier to graph by hand? how? Thanks!

$f(x) = log_2x$

$f(x) =\sqrt{a^2-x^2}$

the second one

$y=\sqrt{a^2-x^2}$ it is the upper semi circle of the circle $y^2+x^2=a^2$ intersect with the x-axis at the points x=a and x=-a
August 7th 2009, 05:26 PM
live_laugh_luv27

Quote:

Originally Posted by Amer

the second one

$y=\sqrt{a^2-x^2}$ it is the upper semi circle of the circle $y^2+x^2=a^2$ intersect with the x-axis at the points x=a and x=-a

ok..How would I set up the axes?
August 7th 2009, 11:00 PM
Amer

1 Attachment(s)

Quote:

Originally Posted by live_laugh_luv27

ok..How would I set up the axes?

like this

Attachment 12390
August 9th 2009, 09:32 AM
live_laugh_luv27

I'm still confused about this graph...http://www.mathhelpforum.com/math-he...60aa7cfc-1.gif

I need to find domain, range, even/odd etc.
August 9th 2009, 09:51 AM
Amer

Quote:

Originally Posted by live_laugh_luv27

I'm still confused about this graph...http://www.mathhelpforum.com/math-he...60aa7cfc-1.gif

I need to find domain, range, even/odd etc.

Treat a like a real number

Domain $0 \leq a^2-x^2 \Rightarrow -a\leq x\leq a$

Range $x^2+y^2 \leq a^2$ with $0 \leq y \leq a$

f(x)=f(-x) so the function is even
August 9th 2009, 10:13 AM
live_laugh_luv27

Quote:

Originally Posted by Amer

Treat a like a real number

Domain $0 \leq a^2-x^2 \Rightarrow -a\leq x\leq a$

Range $x^2+y^2 \leq a^2$ with $0 \leq y \leq a$

f(x)=f(-x) so the function is even

wow, can't believe I didn't see that. Thanks!