NZ Level 7 (NZC) Level 2 (NCEA) Find the Equation of a Square Root Function
Lesson

We now turn to the task of finding the equation of a root function given its graph.

We specifically focus on the transformed basic function given as $y=a\sqrt{x-h}+k$y=axh+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.

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=ax3+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=ax3+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=3x3+2, which is shown in the above graph. Notice that it does indeed passes through $\left(7,-4\right)$(7,4)

## Example 2:

Consider the function of the form $y=a\sqrt{-x}+k$y=ax+k shown below. Can we identify the specific graph from the sketch? 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=ax+10, and all we have to do is determine the size of the dilation.

Three points have been circled on the graph. These look to be, by observation, points that lie on the graph. However looks can be deceiving! We will attempt to find a using all three points, one at at time.

If we select $\left(-16,0\right)$(16,0) then we can write $0=a\sqrt{-\left(-16\right)}+10$0=a(16)+10 and solve for $a$a. The expression simplifies to $4a=-10$4a=10 and so $a=-\frac{5}{2}$a=52 or as a decimal $a=-2.5$a=2.5.

If we select $\left(-10,2\right)$(10,2) then we can write $2=a\sqrt{-\left(-10\right)}+10$2=a(10)+10. This slightly more difficult equation can be solved as follows:

 $2$2 $=$= $a\sqrt{-\left(-10\right)}+10$a√−(−10)+10 $2$2 $=$= $a\sqrt{10}+10$a√10+10 $-8$−8 $=$=  $a$a $=$= $-\frac{8}{\sqrt{10}}$−8√10​ $a$a $=$= $-2.5298...$−2.5298...

This is a worrying result, because it disagrees with our first derived value of $a$a. A quick look back at the graph puts doubt in our mind that the second point $\left(-10,2\right)$(10,2) perhaps might just miss the graph. This is a great reminder that forming equations from a graph has its problems.

We need some confirmation on our first answer of $a=-2.5$a=2.5.

If we select the final point $\left(-4,5\right)$(4,5) then we can write $5=a\sqrt{-\left(-4\right)}+10$5=a(4)+10 and solve for a. The expression easily simplifies to $2a=-5$2a=5 and so $a=-\frac{5}{2}$a=52 or as a decimal $a=-2.5$a=2.5.

Thus, with some sureness we can conclude with reasonable confidence that the specific graph we are looking for is given by $y=-\frac{5}{2}\sqrt{-x}+10$y=52x+10.

#### More Worked Examples

##### QUESTION 1

The square root function that has been graphed has an equation of the form $y=a\sqrt{x}$y=ax.

1. By using a point on the graph, solve for the value of $a$a.

2. Hence state the equation of the function.

##### QUESTION 2

The square root function that has been graphed has an equation of the form $y=a\sqrt{x}+b$y=ax+b.

1. State the value of $b$b.

2. By using a point on the graph, solve for the value of $a$a.

3. Hence state the equation of the function.

##### QUESTION 3

Consider the square root function that has been graphed.

1. For a square root function whose equation is of the form $y=a\sqrt{x-h}+k$y=axh+k, which of the following are the coordinates of the endpoint (where the function is not continuous)?

$\left(a,k\right)$(a,k)

A

$\left(k,a\right)$(k,a)

B

$\left(k,h\right)$(k,h)

C

$\left(h,k\right)$(h,k)

D

$\left(a,k\right)$(a,k)

A

$\left(k,a\right)$(k,a)

B

$\left(k,h\right)$(k,h)

C

$\left(h,k\right)$(h,k)

D
2. State the coordinates of the endpoint of the function that has been graphed.

3. Using a point on the curve, solve for the value of $a$a.

4. Hence state the equation of the square root 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