# graphing

• Nov 30th 2008, 07:11 PM
william
graphing
How do I graph
a)y=2x^2-8x+10

b)y=3x^2+12x-7 and it's inverse, and state the domain and range of each.

• Nov 30th 2008, 07:19 PM
Chris L T521
Quote:

Originally Posted by william
How do I graph
a)y=2x^2-8x+10

and it's inverse, and state the domain and range of each.

Here's the graph to part (a):

http://img.photobucket.com/albums/v4..._inverses1.jpg

Can you find the domain and range of the original function (in blue) and the inverse (in cyan)??
• Nov 30th 2008, 07:22 PM
william
Quote:

Originally Posted by Chris L T521
Here's the graph to part (a):

http://img.photobucket.com/albums/v4..._inverses1.jpg

Can you find the domain and range of the original function (in blue) and the inverse (in cyan)??

d: xEr
r: y>2

Also, how would I go abouts graphing this by hand?
• Nov 30th 2008, 07:27 PM
Chris L T521
Quote:

Originally Posted by william
d: xEr
r: y>2

Correct. Note that range is actually $y\geq 2$ Now what is the domain and range of the inverse?

Quote:

Also, how would I go abouts graphing this by hand?
Pick x values (i.e. x=-2,1,0,1,2,3,4) and evaluate the function at these values. Then plot these points and connect them (make sure the curve looks smooth).
• Nov 30th 2008, 07:37 PM
william
Quote:

Originally Posted by Chris L T521
Correct. Note that range is actually $y\geq 2$ Now what is the domain and range of the inverse?
I'm not too sure. I believe it has the same vertex, so is it the same?

Pick x values (i.e. x=-2,1,0,1,2,3,4) and evaluate the function at these values. Then plot these points and connect them (make sure the curve looks smooth).

I believe it has the same vertex, so would it be the same?Thanks. Would this method be appropriate to use for the second questions as well? Also, for the graphing the inverse by hand I would simply switch the x and y values, correct?
• Nov 30th 2008, 08:20 PM
Chris L T521
Quote:

Originally Posted by william
I believe it has the same vertex, so would it be the same?

Yes, it seems that they have the same vertex.

Quote:

Thanks. Would this method be appropriate to use for the second questions as well?
Yes.

Quote:

Also, for the graphing the inverse by hand I would simply switch the x and y values, correct?
Yes. This time, look where y=-2,-1,0,1,2,3,4. Then connect the points.

Take note that the domain and the range switch for the inverse.