Graph square root of 48y=x

$\sqrt{48 y} = x$

Rewrite the equation as $x = \sqrt{48 y}$ .

$x = \sqrt{48 y}$

Simplify $\sqrt{48 y}$ .

Rewrite $48 y$ as ${(2^{2})}^{2} \cdot (3 y)$ .

Factor $16$ out of $48$ .

$x = \sqrt{16 (3) y}$

Rewrite $16$ as $4^{2}$ .

$x = \sqrt{4^{2} \cdot 3 y}$

Rewrite $4$ as $2^{2}$ .

$x = \sqrt{{(2^{2})}^{2} \cdot 3 y}$

Add parentheses.

$x = \sqrt{{(2^{2})}^{2} \cdot (3 y)}$

Pull terms out from under the radical.

$x = 2^{2} \sqrt{3 y}$

Raise $2$ to the power of $2$ .

$x = 4 \sqrt{3 y}$

The domain in terms of $y$ are all the $y$ values that make the radicand non-negative.

$[0, \infty)$

${y | y \geq 0}$

To find the radical expression end point, substitute the $y$ value $0$ , which is the least value in the domain, into $f (y) = 4 \sqrt{3 y}$ .

Replace the variable $y$ with $0$ in the expression.

$f (0) = 4 \sqrt{3 (0)}$

Simplify the result.

Multiply $3$ by $0$ .

$f (0) = 4 \sqrt{0}$

Rewrite $0$ as $0^{2}$ .

$f (0) = 4 \sqrt{0^{2}}$

Multiply by zero.

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = 4 \cdot 0$

Multiply $4$ by $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

The radical expression end point is $(0, 0)$ .

$(0, 0)$

Select a few $y$ values from the domain. It would be more useful to select the values so that they are next to the $y$ value of the radical expression end point.

Substitute the $y$ value $1$ into $4 \sqrt{3 y}$ . In this case, the point is $(6.93, 1)$ .

Replace the variable $y$ with $1$ in the expression.

$f (1) = 4 \sqrt{3 (1)}$

Simplify the result.

Multiply $3$ by $1$ .

$f (1) = 4 \sqrt{3}$

The final answer is $4 \sqrt{3}$ .

$y = 4 \sqrt{3}$

Substitute the $y$ value $2$ into $4 \sqrt{3 y}$ . In this case, the point is $(9.8, 2)$ .

Replace the variable $y$ with $2$ in the expression.

$f (2) = 4 \sqrt{3 (2)}$

Simplify the result.

Multiply $3$ by $2$ .

$f (2) = 4 \sqrt{6}$

The final answer is $4 \sqrt{6}$ .

$y = 4 \sqrt{6}$

The square root can be graphed using the points around the vertex $(0, 0), (6.93, 1), (9.8, 2)$

$\begin{array}{c} x & y \\ 0 & 0 \\ 6.93 & 1 \\ 9.8 & 2 \end{array}$

Graph square root of 48y=x

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