Lesson

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.

There are a number of theories about the origin of the square root symbol. One of these suggests that, in the early 16th century, Christoph Rudolff, the author of the first German text book on Algebra, introduced the symbol as the lower case letter $r$`r` standing for radix, radical or as we more commonly say root.

Although no geometric square exists with zero area, we still define $\sqrt{0}$√0 as being a valid element of the range of the function.

Negative values of $x$`x` are excluded from the domain when we consider the square root function over the real numbers. Thus the domain is given formally as $\left\{x:x\in\Re^+\cup(x=0)\right\}${`x`:`x`∈ℜ+∪(`x`=0)}, so that the *argument* $x$`x` can be any non-negative real number.

The curve of the graph of $y=\sqrt{x}$`y`=√`x` is one that begins at the origin and rises in a concave downward way, so that the rate of rise is forever slowing although never stopping to rise.

$x$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 |

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.

This gives rise to the square root function's recognisable shape when graphed.

Of course square root functions can have a multitude of arguments. For example $y=\sqrt{x-5}$`y`=√`x`−5, $y=\sqrt{3-2x}$`y`=√3−2`x`, $y=\sqrt{1-x^2}$`y`=√1−`x`2 and $y=2\sqrt{\frac{1+x}{1-x}}$`y`=2√1+`x`1−`x` can all be considered square root functions.

Of course, as the arguments become more and more complex, the determination of the restrictions on the domain and range become more and more difficult.

For example, to determine the domain of the function $y=\sqrt{3-x}$`y`=√3−`x`, we need to ensure that the argument $3-x$3−`x` never becomes negative. So we set $3-x\ge0$3−`x`≥0 and solve for $x$`x`. Adding $x$`x` to both sides and reversing the sense of the inequality reveals $x\le3$`x`≤3 as the correct domain.

The range is governed by the domain. At $x=3$, $y=\sqrt{3-3}=0$ and as x decreases, y increases monotonically toward infinity (although at an ever slowing rate). Hence the range includes all non-negative reals. Formally this means $\left\{y:y\in\Re^+\cup(y=0)\right\}${`y`:`y`∈ℜ+∪(`y`=0)}.

A function like $y=\sqrt{1-x^2}$`y`=√1−`x`2 is interesting in that it can be looked at in more than one way.

The argument is again restricted so that $1-x^2\ge0$1−`x`2≥0, and a little thought should convince you that the choice of $x$`x` values is confined to numbers whose absolute value is less than or equal to $1$1.

At $x=1$`x`=1, $y=\sqrt{1-1^2}=0$`y`=√1−12=0 and at $x=-1$`x`=−1, $y=\sqrt{1-\left(-1\right)^2}=0$`y`=√1−(−1)2=0. The maximum value of $y$`y` clearly occurs when $x=0$`x`=0 when $y$`y` becomes $y=\sqrt{1-0^2}=1$`y`=√1−02=1. Therefore the range includes all reals between and including $0$0 and $1$1.

Taking square roots of numbers between $0$0 and $1$1 (but not including $0$0 or $1$1) increases their size. For example $\sqrt{0.25}=0.5$√0.25=0.5 and $\sqrt{0.81}=0.9$√0.81=0.9 and so on.

Now think of another function - the function $y=1-x^2$`y`=1−`x`2 as the parabola that is inverted and translated up $1$1 unit. If we restrict the domain of this new curve to numbers between $-1$−1 and $1$1 we end up with the graph shown below as the dashed curve.

Thus we can think of the function $y=\sqrt{1-x^2}$`y`=√1−`x`2 as the function whose ordinates (y values) are the square root of the ordinates of $y=1-x^2$`y`=1−`x`2. Each ordinate is extended a little.

The curve given by $y=\sqrt{1-x^2}$`y`=√1−`x`2 is a *semicircle* centred on the origin as shown.

Alternatively, we know that the circle of radius $1$1, centred on the origin, is defined as the locus of the set of points $1$1 unit from the origin. This means that the distance between the origin and the variable point $\left(x,y\right)$(`x`,`y`) is held constant at $1$1 unit.

By the distance formula, we simply set $\sqrt{\left(x-0\right)^2+\left(y-0\right)^2}=1$√(`x`−0)2+(`y`−0)2=1, and this, after squaring both sides, simplifies easily to $x^2+y^2=1$`x`2+`y`2=1.

By making $y$`y` the subject of the formula, we arrive at $y=\pm\sqrt{1-x^2}$`y`=±√1−`x`2 and the semicircle we need is the one that sits above the $x$`x` axis.

Consider the function $y=\sqrt{x}$`y`=√`x`.

Complete the table of values.

Round any values to two decimal places if necessary.

$x$ `x`$0$0 $1$1 $2$2 $3$3 $4$4 $5$5 $9$9 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ The graph of $y=\sqrt{x}$

`y`=√`x`is given.Loading Graph...Is $y=\sqrt{x}$

`y`=√`x`an increasing function or a decreasing function?Increasing

ADecreasing

BIncreasing

ADecreasing

B

Consider the graph of $y=\sqrt{-x}$`y`=√−`x`.

Loading Graph...

Which single option below gives us the correct domain of the function?

The function is defined for $x>0$

`x`>0.AThe function is defined for $x\le0$

`x`≤0.BThe function is defined for $x\ge0$

`x`≥0.CThe function is defined for $x<0$

`x`<0.DThe function is defined for $x>0$

`x`>0.AThe function is defined for $x\le0$

`x`≤0.BThe function is defined for $x\ge0$

`x`≥0.CThe function is defined for $x<0$

`x`<0.DWhich of the following is true of the range?

Select all the correct options.

$y$

`y`can never be positive, regardless of the value of $x$`x`.A$y$

`y`can never be negative, regardless of the value of $x$`x`.BThe smallest possible value of $y$

`y`is $0$0.C$y$

`y`will sometimes be positive and sometimes be negative.D$y$

`y`cannot equal $0$0.E$y$

`y`can never be positive, regardless of the value of $x$`x`.A$y$

`y`can never be negative, regardless of the value of $x$`x`.BThe smallest possible value of $y$

`y`is $0$0.C$y$

`y`will sometimes be positive and sometimes be negative.D$y$

`y`cannot equal $0$0.E

Consider the function $y=\sqrt{x-3}+2$`y`=√`x`−3+2.

State the domain of the function in the form of an inequality.

State the range of the function.

Which of the following is the graph of $y=\sqrt{x-3}+2$

`y`=√`x`−3+2?Loading Graph...ALoading Graph...BLoading Graph...CLoading Graph...DLoading Graph...ALoading Graph...BLoading Graph...CLoading Graph...D

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

Apply graphical methods in solving problems