Question #fbdc5

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Answer 1 · 2017-04-25T16:05:11

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

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

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

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

We can do $0$ $0$ .

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

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

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

$\sqrt{0}$ $sqrt0$

$0$ $0$

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

Let's try $2$ $2$ as well.

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

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

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

$\sqrt{180}$ $sqrt(180)$

$\approx 13.416$ $~~13.416$

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

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

Do not draw the part of the graph with negative $x$ $x$ 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 $x$ $x$ .

$x \geq 0$ $x>=0$