Help with math hw D:

Hi 0-o I need help with another problem.......I was wondering how to graph equations such as this:
y=-3 for -1 <(or equal to) x < (or equal to) 1

and also

(x-4)^2 + y^2 = 1 for x>31/8

P.S. Sorry if this is the wrong section

Please someone help me D: It's worth a lot of points and I don't understand how to do it =x

Hi 0-o I need help with another problem.......I was wondering how to graph equations such as this:
y=-3 for -1 <(or equal to) x < (or equal to) 1

Graph the equation y = -3. If you can't understand that, think about any value of x, then think about what the equation equals to.

Then "hide" the graph for any value of x that lies outside of the parameters, in your case, -1 <= x <= 1.

(x-4)^2 + y^2 = 1 for x>31/8

This one is a lot more complicated.

Where I would start is separate the y and x variables, so that they're on the opposite sides of the equal sign. Then figure out for which values of x give you the equation y^2 = 0. Just in case you're wondering, x = 4 does not give you y^2 = 0.

Also, beware that since y is squared, any negative value of y will also work to satisfy any positive value of y. That is, y^2 = (-y)^2.

Where I would start is separate the y and x variables, so that they're on the opposite sides of the equal sign.

Or just recognize the general form of a circle... which I hope he knows, if he's going to graph one.

(x-h)^2 + (y-k)^2 = r^2
Equation of a circle centered at point (h,k) with radius r

As for the x>31/8 are you sure? None of the graph has any point >31/8.