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

Jul 23, 2018

see explanation

#### Explanation:

Consider $\sqrt{x} \text{ where } x > 0$

Set $y = \sqrt{x}$ where $x > 0 \implies x \ne 0$

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:
$y = \pm \sqrt{x}$

This is a roundabout way of writing ${\left(\pm y\right)}^{2} = x \text{ where } y \ne 0$

So really this is a quadratic in $y$ which is of the shape $\cup$ but 'rotated' ${90}^{o}$ to the right.

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Consider: $y = - \frac{1}{2} x + 3$ where $x < 0 \implies x \ne 0$

Firstly this is an equation of a straight line: form $y = m x + c$
The gradient ($m \to - \frac{1}{2}$) 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 $y = \frac{1}{2} x + 3 \textcolor{w h i t e}{\text{ddd") ->color(white)("ddd}} y = - \frac{1}{2} \left(- 10\right) + 3 = + 8$

Mark the point $\left(x , y\right) = \left(- 10 , 8\right)$
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 $x \ge 0$