Algebra Section 2.1 Flashcards

Is the relation {(3,3),(5,5),(9,9)} a function?

What is the domain of the relation {(3,3),(5,5),(9,9)}?

What is the range of the relation {(3,3),(5,5),(9,9)}?

Is the relation {(2,5),(2,6),(5,5),(5,6)} a function?

What is the domain of the relation {(2,5),(2,6),(5,5),(5,6)}?

What is the range of the relation {(2,5),(2,6),(5,5),(5,6)}?

How do you determine whether the equation x+y=46 defines y as a function of x?

Does the equation x+y=4 define y as a function of x?

Does the equation y=sqrt(x+31) define y as a function of x?

Evaluate f(x)=9x−6 at x=2.

Evaluate f(x)=9x−6 at x=x+5.

Evaluate f(x)=9x−6 at x=-x.

Evaluate h(x)=x^4 -4x^2 +7 at x=3.

Evaluate h(x)=x^4 -4x^2 +7 at x=-1.

Evaluate h(x)=x^4 -4x^2 +7 at x=-x.

Evaluate h(x)=x^4 -4x^2 +7 at x=3a.

A relation is a function if each input (domain) corresponds to exactly one output (range).
Example of a function: {(3,3), (5,5), (9,9)} has a domain of {3, 5, 9} and a range of {3, 5, 9}.
Example of a non-function: {(2,5), (2,6), (5,5), (5,6)} because the input 2 produces two different outputs.

To determine if an equation defines y as a function of x, rearrange it to solve for y.
If a specific x value can produce multiple y values, the equation is not a function.
The equation x + y = 4 is a function because it gives one unique y for each x.
The equation y = √(x + 31) is a function since it’s explicitly solved for y, yielding one y for each x input.

To evaluate a function at specific values, substitute the values into the function expression and simplify.
For f(x) = 9x - 6:
- f(2) = 12
- f(x + 5) = 9x + 39
- f(-x) = -9x - 6
For h(x) = x⁴ - 4x² + 7:
- h(3) evaluates to 52
- h(-1) results in 4
- h(-x) remains as x⁴ - 4x² + 7, as it doesn’t change with -x input.