3.03 Graphs of square root functions

In the last chapter, we saw that we can rearrange $y^2=x$y2=x so that $y$y is the subject and obtain $y=\pm\sqrt{x}$y=±√x.

We can break this up into two separate functions: $y=\sqrt{x}$y=√x and $y=-\sqrt{x}$y=−√x

Shown below we can see that, since the square root function is not defined for negative values, we have a domain of $x\ge0$x≥0.

The graph of $y=\sqrt{x}$y=√x

The graph of $y=-\sqrt{x}$y=−√x

Let's take a closer look at the function $y=\sqrt{x}$y=√x.

$x$x	$0$0	$1$1	$4$4	$9$9	$16$16	$25$25
$y=\sqrt{x}$y=√x	$0$0	$1$1	$2$2	$3$3	$4$4	$5$5

Notice that in order for the $y$y-values to reach the next integer, the $x$x-values get further and further apart. This gives rise to the square root function's recognisable shape when graphed.

 

Transformations of $y=\sqrt{x}$y=√x

The basic root function can be dilated and translated in a similar way to other functions.

The root function $y=\sqrt{x}$y=√x can be transformed to $y=a\sqrt{x-h}+k$y=a√x−h+k. Use the applet below to experiment with these transformations, and try to summarise your findings. Take careful note of the 'starting point' of the function, as well as the domain and range.

Created with Geogebra

Summary:

• If $a$a is negative, the basic curve is reflected across the $x$x-axis
• The graph is dilated vertically by a factor of $a$a
• The graph is translated horizontally by $h$h units
• The graph is translated vertically by $k$k units
• The vertex point, originally at the origin, is moved to the point $\left(h,k\right)$(h,k)
• The domain of the graph becomes $x\ge h$x≥h. Since the function is only defined when the term under the square root is positive, we need $x-h\ge0$x−h≥0 and so $x\ge h$x≥h.

• The range of the graph becomes $y\ge k$y≥k for $a>0$a>0, or $y\le k$y≤k for $a<0$a<0, due to the vertical shift of $k$k units

 

Horizontal reflection

In the previous chapter, we saw parabolas that opened to the left. We can also reflect the square root function across the $y$y-axis.

This is achieved by the function $y=\sqrt{-x}$y=√−x. Think about what values of $x$x this function is defined for - the square root will return real values when $x$x is negative or zero, and so the domain is $x\le0$x≤0. This is a reflection of the function $y=\sqrt{x}$y=√x about the $y$y-axis.

We can reflect our general form $y=a\sqrt{x-h}+k$y=a√x−h+k by multiplying the expression under the square root by $-1$−1. Doing so, we obtain the form $y=a\sqrt{-\left(x-h\right)}+k$y=a√−(x−h)+k, which can also be written as $y=a\sqrt{h-x}+k$y=a√h−x+k.

 

Worked example

example 1

Consider the function given by $f\left(x\right)=-2\sqrt{3-x}$f(x)=−2√3−x .

(a) State the transformations needed to obtain this function from the basic graph $y=\sqrt{x}$y=√x

Think: We can rewrite this function as $f(x)=-2\sqrt{-\left(x-3\right)}$f(x)=−2√−(x−3). 

Do: The transformations are

• Dilating the graph by a factor of $2$2 from the $x$x-axis (this means $y$y-values are doubled)
• Reflecting the graph across the $x$x-axis (this means multiply the $y$y-values by $-1$−1)
• Reflecting the graph across the $y$y-axis (this means multiply the $x$x-values by $-1$−1)
• Translating the graph $3$3 units to the right (this means add $3$3 to the $x$x-values)

(b) Determine the coordinates of the vertex.

Think: Start with the vertex ($0$0 , $0$0) and apply the above transformations in order:

Do:

($0$0, $0$0) →	($0$0, $2\times0$2×0) = ($0$0, $0$0)	→	( $0$0, $-1\times0$−1×0) = ($0$0, $0$0)	→	($-1\times0$−1×0, $0$0) = ($0$0, $0$0)	→	($3+0$3+0, $0$0) = ($3$3, $0$0)
	Double $y$y-value	→	multiply $y$y-value by $-1$−1	→	multiply $x$x value by $-1$−1	→	add $3$3 to $x$x

The vertex will be at the point $\left(3,0\right)$(3,0) 

(c) State the domain of the function

Think: We need to ensure that $3-x\ge0$3−x≥0.

Do: Rearrange to find the domain will be all real numbers $x$x where $x\le3$x≤3

(d) Determine the intercepts on the axes

Think: Intercepts occur when $x=0$x=0 and when $y=0$y=0

Do: When $x=3$x=3, $y=-2\times\sqrt{0}=0$y=−2×√0=0, and when $x=0$x=0, $y=-2\sqrt{3-0}=-2\sqrt{3}$y=−2√3−0=−2√3. This means that the $x$x- and $y$y-intercepts are the points $\left(3,0\right)$(3,0) and $\left(0,-2\sqrt{3}\right)$(0,−2√3).

(e) Use the information from parts a to d, plus some extra points, to sketch the graph

Think: We have enough information to sketch the graph. Before we do so, however, we can increase the accuracy of the sketch by evaluating a few carefully chosen points - say the points at $x=-1$x=−1, $x=-6$x=−6 and $x=-13$x=−13, since these values create square numbers for the term $3-x$3−x.

Do: Doing so, we have:

$f\left(-1\right)$f(−1)	$=$=	$-2\sqrt{3-\left(-1\right)}=-4$−2√3−(−1)=−4
$f\left(-6\right)$f(−6)	$=$=	$-2\sqrt{3-\left(-6\right)}=-6$−2√3−(−6)=−6
$f\left(-13\right)$f(−13)	$=$=	$-2\sqrt{3-\left(-13\right)}=-8$−2√3−(−13)=−8

So we know that the curve passes through the points $\left(-1,-4\right)$(−1,−4), $\left(-6,-6\right)$(−6,−6) and $\left(-13,-8\right)$(−13,−8).

Putting all of this information together, we can make an accurate sketch of $f\left(x\right)=-2\sqrt{3-x}$f(x)=−2√3−x:

Reflect: Look at the graph that we sketched, and compare it to the graph of $y=\sqrt{x}$y=√x. Can you see the effects of each of the transformations that occurred?

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3

 

Finding the equation from a graph

Strategies

• Determine if the graph opens left or right and find the coordinates of the vertex ($x$x, $y$y).
• Substitute $x=h$x=h and $y=k$y=k into the general form for the left or right opening graph from the box above.
• Use the $y$y-intercept or another point on the graph to solve for $a$a.

 

Worked example

example 2

Determine the equation of the graph below.

Think: The graph opens left with vertex ($4$4, $2$2). Substitute $h=4$h=4, $k=2$k=2 into the general form for a left opening graph.

Do: 

$y$y	$=$=	$a\sqrt{h-x}+k$a√h−x+k	General form for left opening square root graph
$y$y	$=$=	$a\sqrt{4-x}+2$a√4−x+2	Substitute $h=4$h=4 and $k=2$k=2
$6$6	$=$=	$a\sqrt{4-0}+2$a√4−0+2	Use ($0$0, $6$6) to solve for $a$a
$6$6	$=$=	$2a+2$2a+2
$4$4	$=$=	$2a$2a
$a$a	$=$=	$2$2

 

Therefore the function is $y=2\sqrt{4-x}+2$y=2√4−x+2

Practice questions

question 4