6) f(x) = x^2 - 2 a: f(x) = |x| - 1 - 2 b: f(x) = |x| - 1 - 2 c: f(x) = |x| + 1 + 2 d: f(x) = √(x + 1) - 2
Added by Timothy F.
Close
Step 1
The graph is a transformation of the absolute value function, which is f(x) = |x|. Now, let's analyze the given options: a) f(x) = |x+1| - 2 b) f(x) = |x-1| - 2 c) f(x) = |x+1| + 2 Show more…
Show all steps
Your feedback will help us improve your experience
Suzanne W. and 65 other Precalculus educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Let $f(x)=6^{x}$ a. Find $f(-2), f(-1), f(0), f(1),$ and $f(2)$ b. Graph $y=f(x)$
Exponential and Logarithmic Functions and Applications
Logarithmic Functions
In Problems $61-64,$ the graph of a function $f$ is illustrated. Use the graph of fas the first step toward graphing each of the following functions: (a) $F(x)=f(x)+3$ (b) $G(x)=f(x+2)$ (c) $P(x)=-f(x)$ (d) $H(x)=f(x+1)-2$ (e) $Q(x)=\frac{1}{2} f(x)$ (f) $g(x)=f(-x)$ (g) $h(x)=f(2 x)$
Functions and Their Graphs
Graphing Techniques: Transformations
$61-62$ . The graph of $y=f(x)$ is given. Match each equation with its graph. $$ \begin{array}{ll}{\text { (a) } y=f(x-4)} & {\text { (b) } y=f(x)+3} \\ {\text { (c) } y=2 f(x+6)} & {\text { (d) } y=-f(2 x)}\end{array} $$
Functions
Transformations of Functions
Recommended Textbooks
Precalculus with Limits
Precalculus
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD