function graphs

• Aug 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!

$\displaystyle f(x) = log_2x$

$\displaystyle f(x) =\sqrt{a^2-x^2}$
• Aug 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!

$\displaystyle f(x) = log_2x$

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

The first one can be put in as $\displaystyle \frac{\log{x}}{\log{2}}$ 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 $\displaystyle \geq 0$ 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)
• Aug 7th 2009, 11:01 AM
dhiab
Quote:

Originally Posted by artvandalay11
The first one can be put in as $\displaystyle \frac{\log{x}}{\log{2}}$ 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 $\displaystyle \geq 0$ 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)
• Aug 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!

$\displaystyle f(x) = log_2x$

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

the second one

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

Originally Posted by Amer
the second one

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

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

Originally Posted by live_laugh_luv27
ok..How would I set up the axes?

like this

Attachment 12390
• Aug 9th 2009, 09:32 AM
live_laugh_luv27

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

Originally Posted by live_laugh_luv27

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

Treat a like a real number

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

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

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

Originally Posted by Amer
Treat a like a real number

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

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

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

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