Q1. - Algebra 1

Question

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

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

 

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

 

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

 

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

Step-by-Step Solution

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 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.

$k$ 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.