McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
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Exercise 32 Page 440

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

Practice makes perfect

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

Graphing the Function

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.

Finding the Domain and Range

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.

f(x)=-sqrt(6x)
f( 0)=- sqrt(6( 0))
f(0)=0

This tells us that the range is all values of f(x) such that f(x) ≤ 0. Domain:& x ≥ 0 Range:& f(x)≤ 0