Answer:
Part 1
a = [tex]- 1[/tex]
Reflects f(x) over the x-axis
h = [tex]2[/tex]
Translates the graph 2 units to the left horizontally
v = [tex]3[/tex]
Translates the graph 3 units up vertically
Part 2
[tex]x \quad\quad\quad f(x)= -\sqrt{x + 2} + 3\\--------------------------\\\\-2 \quad\quad\quad-\sqrt{-2 + 2} + 3 =- \sqrt{0} + 3 =3 \\\\2 \quad\quad\quad-\;\sqrt{2 + 2} + 3 =- \sqrt{4} + 3 = -2 + 3 = 1\\\\7 \quad\quad\quad-\;\sqrt{7 + 2} + 3 =- \sqrt{9} + 3 = -3 + 3 = 0\\ \\14 \quad\quad\quad -\sqrt{ 14 + 2} + 3 = -\sqrt{16} + 3 = -4 + 3 = -1\\\\23 \quad\quad\quad-\sqrt{23 + 2} + 3 =- \sqrt{5} + 3 =-5 \\\\[/tex]
part 3
See attached graph and explanation
Part 4
Domain: [tex][-2,\:\infty \:)[/tex]
Range: [tex](-\infty \:,\:3][/tex]
Step-by-step explanation:
Part 1
- The given function is:
[tex]f(x)= -\sqrt{x + 2} + 3[/tex]
- The parent function is:
[tex]p(x) = \sqrt{x}[/tex]
- When we prefix a negative sign in front of the previous function we are reflecting the graph over the x-axis.
[tex]g(x) = - p(x) = - \sqrt{x}[/tex]
We are multiplying f(x) by - 1 so a = - 1
- When we add a constant to the term under the square root we shift the original graph two units to the left. This indicates a translation of 2 units horizontally to the left
[tex]h(x) = g(x + 2) = -\sqrt{x+2}[/tex]
The horizontal shift h = +2
- Finally we are adding3 to h(x) to get
[tex]m(x) = h(x) + 3 \\\\m(x) = - \sqrt{x + 2} + 3[/tex]
This represents a vertical shift upward for [tex]h(x) = - \sqrt{x + 2}[/tex]
v = + 3
Part 2
To generate the table, take some values of x, plug it into [tex]f(x)= -\sqrt{x + 2} + 3[/tex]
and compute the corresponding values. Use a calculator
Note that x has to be ≥ -2 otherwise you will get a negative sign under the square root term and square roots of negative numbers are not real
Choose numbers for x such that x + 2 is a perfect square
That makes it easier to work with. For example, x = 0, x = 2, x = 12 etc
Here are my values with calculations. A little messy but readable
[tex]x \quad\quad\quad f(x)= -\sqrt{x + 2} + 3\\--------------------------\\\\-2 \quad\quad\quad-\sqrt{-2 + 2} + 3 =- \sqrt{0} + 3 =3 \\\\2 \quad\quad\quad-\;\sqrt{2 + 2} + 3 =- \sqrt{4} + 3 = -2 + 3 = 1\\\\7 \quad\quad\quad-\;\sqrt{7 + 2} + 3 =- \sqrt{9} + 3 = -3 + 3 = 0\\ \\14 \quad\quad\quad -\sqrt{ 14 + 2} + 3 = -\sqrt{16} + 3 = -4 + 3 = -1\\\\23 \quad\quad\quad-\sqrt{23 + 2} + 3 =- \sqrt{5} + 3 =-5 \\\\[/tex]
Part 3
- The solid green line is the function [tex]f(x)= -\sqrt{x + 2} + 3[/tex]
- The dotted lines of various colors show the result of each of the transformations described
- You don't need to worry about the dotted line graphs, that is just to provide you an understanding of how each transformation affects the previous transformation. Ignore them
Part 4
Domain of f(x)
- The domain of a function is the set of all input values (x) that result in a defined, real value for f(x)
- As mentioned above x cannot be less than -2 since the term under the square root will be negative and result in a non-real value for f(x)
- There is no upper restriction for x - it can go upto but not include ∞
- Therefore the domain of x is
x ≥ -2
or
-2 ≤ x < ∞
or
[tex][-2,\:\infty \:)[/tex] in interval notation
(I am not sure which format they are asking for; all represent the same thing}
Range of a function
- The range of a function is the range of values of y for the domain of x
- We see that f(x) is a decreasing function the minimum value for y is -∞ since there is no limit to how low f(x) can get.
- The maximum value is at x = -2 and is 3
- So we can state the range of f(x) as
x ≤ 3
or
- ∞ < x ≤ x
or
[tex](-\infty \:,\:3][/tex] in interval notation
