McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Study Guide and Review
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Exercise 34 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≥- 5
Range: f(x) ≥ -3

Practice makes perfect

The given function is a square root function. f(x)=sqrt(x+ 5)- 3 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

Graphing the Function

To graph the given 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 translate the graph 3 units down by subtracting 3 to each of the y-coordinates.

For the last transformation, we will translate the graph 5 unit to the left. To do this, we will subtract 5 to each x-coordinate.

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. x+5 ≥ 0 ⇔ x ≥ - 5 Therefore, the possible values of x are those such that x≥ - 5. By substituting the minimum value of the domain into the function, we can find the minimum value of the range.

f(x)=sqrt(x+5)-3
f( - 5)=sqrt(- 5+5)-3
f(- 5)=0-3
f(- 5)=- 3

This tells us that the range is all values of f(x) such that f(x)≥ - 3. Domain:& x ≥ - 5 Range:& f(x)≥- 3