1. 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.

2. 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):

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

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

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?

4. 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?

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).

5. 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?

6. 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.

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

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.