NZ Level 7 (NZC) Level 2 (NCEA)
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Graphing Square Root Functions
Lesson

Graphing tips

When graphing functions there are a number of useful non-calculus tools that you could consider. These could include:

  • Recognising the basic form of the function
  • Recognising embedded dilations and translations
  • Considering the domain and range
  • Plotting a few points
  • Identifying any symmetrical features
  • Finding the axes intercepts
  • Looking at the behaviour of the function at its extremities 

Relying solely on technology can be hazardous. Technology is a great tool, but it can only ever deliver a plot within a rectangular window that may or may not contain some or all of the graph's essential features.

Moreover, an important reason why we classify types of graphs is that they are useful for modelling physical phenomena - the path of a projectile, or a planet, the rise and fall of demand and supply, radioactive decay, depreciation and investment growth just to name a few.

It is thus vitally important to be able to recognise and select forms appropriate to that real world phenomena. This means developing a good understanding of each of the main forms and learning to utilise a few simple and universal tools to sketch them. 

The above list is not exhaustive, and there is no implied order. In fact only some of the considerations might apply for any given function. 

We will attempt to provide you with the language of mathematical investigation using two examples involving root functions.  

Example 1:

Consider the function given by $f\left(x\right)=-2\sqrt{3-x}$f(x)=23x . 

The basic form of this function is the root function $y=\sqrt{x}$y=x, but clearly there are other complications to consider. 

We need to ensure that the argument $3-x\ge0$3x0, and this means that the function's natural domain includes all real numbers $x$x where $x\le3$x3

We also note that 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=230=23. This means that the $x$x and $y$y intercepts are $\left(3,0\right)$(3,0) and $\left(0,-2\sqrt{3}\right)$(0,23).

As $x$x decreases, from $3$3 to $0$0 and into negative numbers, the quantity $3-x$3x increases, but the square root effect slows that increase down.

The factor of $-2$2 applies a reflective dilation to $\sqrt{3-x}$3x so that each generated $y$y value becomes not only twice as big as it would be if it wasn't there, but changes its sign so that it grows negatively. It does so without any limit to its growth, and so the range must be given by $y:-\inftyy:<y0.

Finally, evaluating a few carefully chosen points, say for $x=-1$x=1,  $x=-6$x=6 and $x=-13$x=13 (to avoid problems with the square root) yields:

$f\left(-1\right)$f(1) $=$= $-2\sqrt{3-\left(-1\right)}=-4$23(1)=4
$f\left(-6\right)$f(6) $=$= $-2\sqrt{3-\left(-6\right)}=-6$23(6)=6
$f\left(-13\right)$f(13) $=$= $-2\sqrt{3-\left(-13\right)}=-8$23(13)=8
     

Hence the curve passes through $\left(-1,-4\right)$(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 informed sketch of $f\left(x\right)=-2\sqrt{3-x}$f(x)=23x.

Example 2:

Consider the function given by $y=\frac{1}{2}\sqrt{4x-5}+3$y=124x5+3.

Using the same observations, we first note that the domain is restricted to values of $x$x that satisfy $4x-5\ge0$4x50. Solving this shows that the domain is defined as $x\ge\frac{5}{4}$x54.

Again the basic function is $y=\sqrt{x}$y=x, but there is a translation of $\frac{5}{4}$54 units to the right, in accordance with the domain we ascertained. We find the translation as follows:

$y$y $=$= $\frac{1}{2}\sqrt{4x-5}+3$124x5+3
  $=$= $\frac{1}{2}\sqrt{4\left(x-\frac{5}{4}\right)}+3$124(x54)+3
  $=$= $\frac{1}{2}\times\sqrt{4}\times\sqrt{\left(x-\frac{5}{4}\right)}+3$12×4×(x54)+3
  $=$= $\frac{1}{2}\times2\times\sqrt{\left(x-\frac{5}{4}\right)}+3$12×2×(x54)+3
  $=$= $\sqrt{\left(x-\frac{5}{4}\right)}+3$(x54)+3
     

This is an interesting example, because at first it looks as though there is a dilation factor of $\frac{1}{2}$12, but in fact there is no dilation of the base root function present (that is the dilation factor is $1$1). 

The algebra reveals that there is a horizontal translation to the right of $\frac{5}{4}$54 units and of course a vertical translation of $3$3 units.

Putting $x=0$x=0 shows $y=\sqrt{0-\frac{5}{4}}+3$y=054+3 and since the number inside the square root is negative, we know that there is no $y$y intercept.

Putting $y=0$y=0 shows $0=\sqrt{x-\frac{5}{4}}+3$0=x54+3, and we see at once that such an equation is impossible. The right hand side is definitely positive for any value of $x$x in the domain. This means that there is no $x$x intercept either. 

From our earlier work on the domain and range, it became clear that the curve had neither an $x$x intercept or $y$y intercept anyway.

We can see that the function was a simple translation, away fro the x and y axes, of the basic root function $y=\sqrt{x}$y=x

To finish our investigation, we might plot another point, although in this case its hard to find an integer y value. At $x=10$x=10 for example $y=\sqrt{\left(10\right)-\frac{5}{4}}+3=5.958..$y=(10)54+3=5.958.. which is very close to $6$6, so the curve passes very close to $\left(10,6\right)$(10,6).

The graph of $y=\frac{1}{2}\sqrt{4x-5}+3$y=124x5+3, or in its simpler form $y=\sqrt{x-\frac{5}{4}}+3$y=x54+3 is shown here.

Worked Examples

Question 1

Consider the function $f\left(x\right)=\sqrt{x}$f(x)=x.

  1. Graph the curve.

    Loading Graph...

  2. What is the domain?

    $x>0$x>0

    A

    $x\ge0$x0

    B

    all real numbers

    C

    $x\le0$x0

    D

    $x>0$x>0

    A

    $x\ge0$x0

    B

    all real numbers

    C

    $x\le0$x0

    D
  3. What is the range?

    $y\le0$y0

    A

    $y\ge0$y0

    B

    all real numbers

    C

    $y>0$y>0

    D

    $y\le0$y0

    A

    $y\ge0$y0

    B

    all real numbers

    C

    $y>0$y>0

    D

Question 2

Consider the function $f\left(x\right)=\sqrt{-x}-1$f(x)=x1.

  1. Graph the function $f\left(x\right)=\sqrt{-x}-1$f(x)=x1.

    Loading Graph...

  2. What is the domain?

    $x\le-1$x1

    A

    $x$x $\le$ $0$0

    B

    $x$x $\ge$ $0$0

    C

    $x\ge-1$x1

    D

    $x\le-1$x1

    A

    $x$x $\le$ $0$0

    B

    $x$x $\ge$ $0$0

    C

    $x\ge-1$x1

    D
  3. What is the range?

    $f\left(x\right)$f(x) $\ge$ $0$0

    A

    $f\left(x\right)$f(x) $\ge$ $-1$1

    B

    $f\left(x\right)$f(x) $\le$ $-1$1

    C

    $f\left(x\right)$f(x) $\le$ $0$0

    D

    $f\left(x\right)$f(x) $\ge$ $0$0

    A

    $f\left(x\right)$f(x) $\ge$ $-1$1

    B

    $f\left(x\right)$f(x) $\le$ $-1$1

    C

    $f\left(x\right)$f(x) $\le$ $0$0

    D

Question 3

The function $y=\sqrt{x}$y=x has been transformed (through translation and/or reflection) into a function that has a domain of $x$x $\le$ $2$2 and a range of

$y$y $\ge$ $1$1. Find the equation of the transformed function.

 

Outcomes

M7-2

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

91257

Apply graphical methods in solving problems

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