Question

How do you graph `sqrtx", if "x>0 and -1/2x+3", if "x≤ 0`?

Answer 1

see explanation

Explanation:

Consider `sqrt(x)" where "x>0`

Set `y=sqrt(x)` where `x>0=> x!=0`

Any number that is negative times its self will give a positive answer.

Any number that is positive times itself will give a positive answer.

So in this case `y` can be either positive or negative so we write:
`y=+-sqrt(x)`

This is a roundabout way of writing `(+-y)^2=x" where "y!=0`

So really this is a quadratic in `y` which is of the shape `uu` but 'rotated' `90^o` to the right.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider: `y=-1/2x+3` where `x<0 =>x!=0`

Firstly this is an equation of a straight line: form `y=mx+c`
The gradient (`m->-1/2`) is negative so the line slopes downward reading left to right on the x-axis.

The constant `c` is of value `+3` so the line crosses the y-axis (y-intercept) at `y=3`. However, it is not permitted for `x` to be greater than or equal to 0 so the actual y-intercept is an 'excluded value'.

The plot well terminate at the y-axis so there is no line on and to the right of the y-axis.

Pick any value for `x`: I* choose `10`

then `y=1/2x+3color(white)("ddd") ->color(white)("ddd") y=-1/2(-10)+3 = +8`

Mark the point `(x,y)=(-10,8)`
Using a ruler line this up with the value of 3 on the x-axis and draw your line making sure you do not go beyond the y-axis into the zone `x>=0`