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What differences do you see between the given function and the parent function? Apply those transformations to the graph of the parent function, f(x)=sqrt(x).
Graph:
Domain: x≥0
Range: f(x) ≤0
The given function is a square root function. The graph of it will be a transformed version of the parent function, y=sqrt(x). Square root functions typically follow the same general format. f(x)= asqrt(x- h)+ k To graph the given square root function, let's put it in general form first. f(x)=-sqrt(6x) ⇔ f(x)= - sqrt(6)sqrt(x- 0)+ 0
Let's put the minus sign aside before we start to graph the given function. We will first graph f(x)=sqrt(6)sqrt(x) and then we will reflect it across the x-axis. To graph the function, let's show the possible transformations of f(x)=sqrt(x).
| Transformations of f(x) | |
|---|---|
| Vertical Translations | Translation up k units, k>0 y=f(x)+ k |
| Translation down k units, k>0 y=f(x)- k | |
| Horizontal Translations | Translation right h units, h>0 y=f(x- h) |
| Translation left h units, h>0 y=f(x+ h) | |
| Vertical Stretch or Shrink | Vertical stretch, a>1 y= af(x) |
| Vertical shrink, 0< a<1 y= af(x) | |
| Reflections | In the x-axis y= - f(x) |
| In the y-axis y= f(- x) | |
Using the table, we can graph the function as a series of transformations. Let's begin with the parent function.
Next, we will multiply the y-coordinates by a= sqrt(6). This stretches the parent graph by a factor of sqrt(6).
Since h and k are zero, neither horizantal nor vertical translation is needed. The last thing to do is to reflect the graph of f(x)=sqrt(6x) across the x-axis because there is a negative sign in front of the given function. To do this, we will put ( -) sign to each y-coordinate.
| Points | Reflection in the x-axis |
|---|---|
| (1,sqrt(6)) | (1, -sqrt(6)) |
| (4,2sqrt(6)) | (4, -2sqrt(6)) |
Finally, we have the graph of the given function.
To determine the domain of the function, recall that the radicand cannot be negative. 6x ≥ 0 ⇔ x ≥ 0 Therefore, the possible values of x are those such that x≥ 0. We have a decreasing function and it always takes negative values. By substituting the minimum value of the domain into the function, we can find the maximum value of the range.
This tells us that the range is all values of f(x) such that f(x) ≤ 0. Domain:& x ≥ 0 Range:& f(x)≤ 0