# Question #fbdc5

Apr 25, 2017

$x \ge 0$
graph{(sqrt(45x^2)) [-6.67, 5.814, -1.274, 4.966]}

#### Explanation:

Assume that variables represent nonnegative real numbers.
So, $x$ can't be negative.

$\sqrt{45 {x}^{2}}$

All we need to do is give $x$ any positive value we want.

We can do $0$.

$\sqrt{45 {x}^{2}}$

$\sqrt{45 {\left(0\right)}^{2}}$

$\sqrt{45 \left(0\right)}$

$\sqrt{0}$

$0$

This means our point on $x$ is $0$
and our point on $y$ is $0$
$\left(0 , 0\right)$

Let's try $2$ as well.

$\sqrt{45 {x}^{2}}$

$\sqrt{45 {\left(2\right)}^{2}}$

$\sqrt{45 \left(4\right)}$

$\sqrt{180}$

$\approx 13.416$

This means our point on $x$ is $2$
and our point on $y$ is $13.416$
$\left(2 , 13.416\right)$

Keep doing this until you have all of the positive $x$ values that you need to draw your graph.

Do not draw the part of the graph with negative $x$ values because that is the rule from our question.

Your graph should look like the answer I gave you, but without the line on the left side of the y-axis. Thos are negative $x$.

$x \ge 0$