Question

What is the equation of the square root function graphed below.

$\mathit{A}$$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\sqrt{\mathbf{x}}\mathbf{+}\mathbf{1}$

$\mathit{B}$$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\sqrt{\mathbf{x}}\mathbf{+}\mathbf{3}$

$\mathit{C}$ $\mathit{y}\mathbf{=}\mathbf{2}\sqrt{\mathbf{x}}\mathbf{+}\mathbf{3}$

$\mathit{D}$$\mathit{y}\mathbf{=}\mathbf{2}\sqrt{\mathbf{x}}\mathbf{+}\mathbf{1}$

Short answer

The equation of the square root function graphed is$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\sqrt{\mathbf{x}}\mathbf{+}\mathbf{3}$.

Option $\mathit{B}$is correct.

Step-by-step solution

Step 1. State the general form of the ‘square root function’.

The general form of a square root function is $y=a\sqrt{x-h}+k$, where

$h$ represents horizontal translation.

If $‘h’$is positive, then the graph $y=\sqrt{x}$shifts h units to the right.

If $‘h’$is negative, then the graph $y=\sqrt{x}$shifts h units to the left.

localid="1650898228847" $k$represents vertical translation.

If $‘k’$is positive, then the graph $y=\sqrt{x}$shifts k units upward.

If $‘k’$is negative, then the graph $y=\sqrt{x}$shifts k units downward.

$‘a’$represents dilation if $‘a’$is positive and $‘a’$represents reflection if $‘a’$is negative.

Note: Here$(h,k)$is the starting endpoint of the curve $y=a\sqrt{x-h}+k$.

Step 2. Calculate the equation of the graph.

Observe the graph given below.

From the graph, notice that the starting endpoint point is $(0,3)$.

Therefore, $(h,k)=(0,3)$where$h=0$ and $k=3$.

$h=0$tells that there is no horizontal translation.

$k=3$tells that the graph is translated up by 3 units.

Also see that the graph passes throught the pont $(1,1)$.

Therefore,$(x,y)=(1,1)$ where$x=1$ and $y=1$.

It can be observed that the function is reflected across the $X$-axis. Therefore $‘a’$must be negative.

Substitute $(h,k)=(0,3)$and $(x,y)=(1,1)$in $y=a\sqrt{x-h}+k$and calculate the value of $‘a’$

$\begin{array}{l}y=a\sqrt{x-h}+k\\ 1=a\left(\sqrt{1-0}\right)+3\\ 1=a\left(\sqrt{1}\right)+3\\ 1=a\left(1\right)+3\\ 1=a+3\\ 1-3=a\\ -2=a\\ a=-2\end{array}$

Substitute $h=0$and $k=3$and $a=-2$in $y=a\sqrt{x-h}+k$to get the required equation.

$\begin{array}{c}y=-2\sqrt{x-0}+3\\ y=-2\sqrt{x}+3\end{array}$

Step 3. State the conclusion.

Therefore, the equation of the graph is$y=-2\sqrt{x}+3$.

Option $B$is correct.