4.05 Graphs and characteristics of square root functions

The parent function $f\left(x\right)=\sqrt{x}$f(x)=√x

The basic square root function is given by $y=\sqrt{x}$y=√x.

The name 'root' derives from an old Scandinavian word meaning the source or origin of something. The Latin form of the word is Radix, and square roots are often called radicals for that reason.

The square root $\sqrt{x}$√x can be interpreted as the side length that is needed to create a square of area $x$x as shown in this diagram.

 

Characteristics of the square root function

Although no geometric square exists with length and area of zero, we still define $\sqrt{0}$√0 as being a valid element of the domain and range of the function. However, negative values of $x$x are excluded from the domain as we can't have a negative length.

Let's work towards graphing $f\left(x\right)=\sqrt{x}$f(x)=√x by starting with a table of values.

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

Note from the table that it takes a jump of $3$3 from $x=1$x=1 to $x=4$x=4 to take $y$y from $1$1 to $2$2. It takes a jump of $5$5 to take $y$y from $2$2 to $3$3 and a jump of $7$7 to take $y$y from $3$3 to $4$4. The pattern continues indefinitely - the size of the increases in $x$x to move $y$y up by $1$1 unit continues to climb. As the $x$x-value increases, the $y$y-value also increases, but the rate of change is decreasing as seen in the graph below.

Using the graph we can identify the key characteristics of the square root function, $f\left(x\right)=\sqrt{x}$f(x)=√x.

Characteristic	$f\left(x\right)=\sqrt{x}$f(x)=√x
Domain	Inequality form: $x\ge0$x≥0 Interval form: $\left[0,\infty\right)$[0,∞)
Range	Inequality form: $y\ge0$y≥0 Interval form: $\left[0,\infty\right)$[0,∞)
Vertex/endpoint/extrema	$\left(0,0\right)$(0,0)
$x$x-intercept	$\left(0,0\right)$(0,0)
$y$y-intercept	$\left(0,0\right)$(0,0)
Increasing/decreasing	Increasing over its domain
End behavior	As $x\to\infty$x→∞, $y\to\infty$y→∞

 

Practice question

Question 1

 

Transformations of $y=\sqrt{x}$y=√x

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

The root function $f(x)=\sqrt{x}$f(x)=√x can be transformed to $g(x)=a\sqrt{b(x-h)}+k$g(x)=a√b(x−h)+k by dilating and/or reflecting it using the factors $a$a and $b$b, translating it, first horizontally $h$h units to the right, and then $k$k units upward.

Notation

Using function notation, if $f\left(x\right)=\sqrt{x}$f(x)=√x then, for example, after a dilation of $2$2, and translations of  $5$5 units to the right and $3$3 units down, we create a new transformed function, say $g\left(x\right)$g(x) ,where:

$g\left(x\right)=2\cdot f\left(x-5\right)-3=2\sqrt{x-5}-3$g(x)=2·f(x−5)−3=2√x−5−3

Exploration

The best thing to do is to experiment with the first applet below showing how the variables $a,h$a,h and $k$k change the basic curve. Try both negative and positive values of $a$a.  

Focusing on the domain and range of the transformed function $y=a\sqrt{x-h}+k$y=a√x−h+k, note that we need $x-h\ge0$x−h≥0 and thus $x\ge h$x≥h.

The new range, because of the translation up or down caused by $k$k, has also changed to the interval $k\le y<\infty$k≤y<∞ for values of $a>0$a>0. If however $a$a is negative the new range becomes $-\infty−∞<y≤k. Experiment with the applet, taking careful note of the natural domains and ranges.

 

Summary

Dilations

$g(x)=a\sqrt{x}$g(x)=a√x is the graph of $f(x)=\sqrt{x}$f(x)=√x dilated vertically $\left(1,1\right)$(1,1) $\to$→ $\left(1,a\right)$(1,a)	If $a>1$a>1, then vertical dilation (stretch) away from the $x$x-axis If$00<a<1, then vertical dilation (compression) toward the $x$x-axis
$g(x)=\sqrt{ax}$g(x)=√ax is the graph of $f(x)=\sqrt{x}$f(x)=√x dilated horizontally $\left(1,1\right)$(1,1) $\to$→ $\left(\frac{1}{a},1\right)$(1a​,1)	If $a>1$a>1, then horizontal dilation (compression) toward the $y$y-axis If $00<a<1, then horizontal dilation (stretch) away from the $y$y-axis

Reflections

$g(x)=-\sqrt{x}$g(x)=−√x is the graph of $f(x)=\sqrt{x}$f(x)=√x reflected vertically $\left(1,1\right)$(1,1) $\to$→ $\left(1,-1\right)$(1,−1)	Reflected over the $x$x-axis Range becomes $y\le0$y≤0
$g(x)=\sqrt{-x}$g(x)=√−x is the graph of $f(x)=\sqrt{x}$f(x)=√x reflected horizontally $\left(1,1\right)$(1,1) $\to$→ $\left(-1,1\right)$(−1,1)	Reflected over the $y$y-axis Domain becomes $x\le0$x≤0

Translations

$g(x)=\sqrt{x}+k$g(x)=√x+k is the graph of $f(x)=\sqrt{x}$f(x)=√x translated vertically $\left(0,0\right)$(0,0) $\to$→ $\left(0,k\right)$(0,k)	If $k>0$k>0, then translated $k$k units up If $k<0$k<0, then translated $k$k units down Range becomes $y\ge k$y≥k
$g(x)=\sqrt{x-h}$g(x)=√x−h is the graph of $f(x)=\sqrt{x}$f(x)=√x translated horizontally $\left(0,0\right)$(0,0) $\to$→ $\left(h,0\right)$(h,0)	If $h>0$h>0, then translated $h$h units right If $h<0$h<0, then translated $h$h units left Domain becomes $x\ge h$x≥h

 

Practice questions

QUESTION 2

QUESTION 3

QUESTION 4

 

Finding the equation given its graph

We specifically focus on the transformed basic function given as $y=a\sqrt{x-h}+k$y=a√x−h+k, which has the three unknown constants $a,h$a,h and $k$k.

As we have found from previous work, the starting point for such a function is the point $\left(h,k\right)$(h,k), a coordinate position completely independent of the dilation factor $a$a.

By independent we mean that changing the dilation factor has no effect on $\left(h,k\right)$(h,k).  The following graph shows various dilated root functions of the form $y=a\sqrt{x-3}+2$y=a√x−3+2. 

What the independence means is that it now becomes straightforward to find the correct dilation factor $a$a once the starting point has been determined.

For example, suppose we established that the root function $y=a\sqrt{x-3}+2$y=a√x−3+2 that starts at the point $\left(3,2\right)$(3,2), also passes through the point $\left(7,-4\right)$(7,−4). 

Since $\left(7,-4\right)$(7,−4) is on the curve, it must satisfy the equation, so we proceed as follows:

$y$y	$=$=	$a\sqrt{x-3}+2$a√x−3+2
$-4$−4	$=$=	$a\sqrt{7-3}+2$a√7−3+2
$-4$−4	$=$=	$a\sqrt{4}+2$a√4+2
$-4$−4	$=$=	$2a+2$2a+2
$-6$−6	$=$=	$2a$2a
$a$a	$=$=	$-3$−3

Therefore the correct equation is given by $y=-3\sqrt{x-3}+2$y=−3√x−3+2, which is shown in the above graph. Notice that it does indeed passes through $\left(7,-4\right)$(7,−4). 

Worked example

Question 5

Consider the function of the form $y=a\sqrt{-x}+k$y=a√−x+k shown below. Can we identify the specific graph from the sketch? 

Think:

From the graph we see that the starting point is $\left(0,10\right)$(0,10) and this explains why the general form given does not show the presence of the horizontal translation $h$h - there is no shift left or right from the origin.

So the function we are dealing with is given by $y=a\sqrt{-x}+10$y=a√−x+10, and all we have to do is determine the size of the dilation. For this we need a point on the curve, we can see that $\left(-16,0\right)$(−16,0) is on the curve.

Do:

$0$0	$=$=	$a\sqrt{-\left(-16\right)}+10$a√−(−16)+10
$0$0	$=$=	$a\sqrt{16}+10$a√16+10
$-10$−10	$=$=	$a\sqrt{16}$a√16
$-10$−10	$=$=	$4a$4a
$a$a	$=$=	$\frac{-10}{4}$−104​
$a$a	$=$=	$-\frac{5}{2}$−52​

The specific graph we are looking for is given by $y=-\frac{5}{2}\sqrt{-x}+10$y=−52​√−x+10.

 

Practice questions

QUESTION 6

QUESTION 7

 

Applications of the square root function

 

Worked example

Question 8

On the Earth, when a rock is dropped from a $100$100 meter cliff face the elapsed time $t$t in seconds after it has fallen $d$d meters will be approximately given by $t_e=0.452\sqrt{d}$te​=0.452√d . On the Moon however, from a $100$100 meter cliff face, the rock would fall so that the elapsed time  after $d$d meters would be given by $t_m=1.11\sqrt{d}$tm​=1.11√d. Compare the elapsed times for the Earth and Moon for a rock falling from $20$20 m and $80$80 m.

Think: Perhaps the best way to answer this question is to graph both functions using technology.

Do: For $20$20 m, we can read off the graph that the Moon rock takes $5$5 seconds, whereas the Earth rock only takes $2$2 seconds.

For $80$80 m, we can read off the graph that the Moon rock takes $10$10 seconds, whereas the Earth rock only takes $4$4 seconds. 

Reflect: It seems that for both the Earth and the Moon it takes twice as long to fall four times the distance. Isaac Newton discovered a universal law of falling 'bodies'. Ignoring air resistance, for any planet, if a body (such as a rock) takes $t$t seconds to fall $d$d meters, then it will take  $kt$kt seconds to fall $k^2d$k2d meters.

This applet will allow you to compare the falling times from any planet (and the moon) with Earth. Remember that when comparing gravitational attractions, its not really about the size of a planet, but rather more about the density of its matter. The gas giant Uranus for example has almost the same fall curve as the Earth, but there is a vast difference between their sizes.  

 

Practice questions

QUESTION 9

QUESTION 10