1. Y = 2x - 1 / (x^2 - x), Domain? 2. f(x) = √(1 / (x - 1)), Domain & Range? 3. Draw the graph of the function f(x) = x^2 & explain how to get the graph of g(x) = (x - 2)^2 from th... 1. Y = 2x - 1 / (x^2 - x), Domain? 2. f(x) = √(1 / (x - 1)), Domain & Range? 3. Draw the graph of the function f(x) = x^2 & explain how to get the graph of g(x) = (x - 2)^2 from the graph of g(x) = x^2. 4. Does the equation, x = |Y| with x = 0, represent Y as a function of x? 5. Draw the graph of |X| + |Y| = 10.
Understand the Problem
The question contains multiple math problems, including functions, domains, ranges, graphing functions, and an equation involving absolute values. It requires solving and drawing related graphs for the given mathematical expressions.
Answer
- Domain of \( y = \frac{2x - 1}{x^2 - x} \): \( x \in \mathbb{R} \setminus
Answer for screen readers
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The domain of ( y = \frac{2x - 1}{x^2 - x} ) is ( x \in \mathbb{R} ) except ( x = 0, 1 ).
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The domain of ( f(x) = \sqrt{\frac{1}{x - 1}} ) is ( (1, \infty) ) with a range of ( (0, \infty) ).
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( f(x) = x^2 ) graphs as a parabola.
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To obtain ( g(x) = (x - 2)^2 ), shift ( f(x) ) right by 2 units.
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( x = |y| ) with ( x = 0 ) implies ( y = 0 ).
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The shape defined by ( |x| + |y| = 10 ) is a diamond centered at the origin with vertices at ( (10, 0) ), ( (0, 10) ), ( (-10, 0) ), ( (0, -10) ).
Steps to Solve
- Finding Domain of the First Function
For the function ( y = \frac{2x - 1}{x^2 - x} ), we need to find the values of ( x ) that do not make the denominator zero.
Solve ( x^2 - x = 0 ):
$$ x(x - 1) = 0 $$
This gives ( x = 0 ) and ( x = 1 ). Thus, the domain is all real numbers except ( x = 0 ) and ( x = 1 ).
- Finding Domain and Range of the Second Function
For the function ( f(x) = \sqrt{\frac{1}{x - 1}} ), we need ( x - 1 > 0 ) for the expression under the square root to be positive.
So,
$$ x > 1 $$
Thus, the domain is ( (1, \infty) ).
To find the range, observe that as ( x ) approaches ( 1 ) from the right, ( f(x) ) approaches ( \infty ). As ( x ) approaches ( \infty ), ( f(x) ) approaches ( 0 ).
Thus, the range is ( (0, \infty) ).
- Graphing ( f(x) = x^2 )
The graph of ( f(x) = x^2 ) is a parabola that opens upwards. It is symmetric about the y-axis and passes through the origin.
Include points like ( (-2, 4) ), ( (-1, 1) ), ( (0, 0) ), ( (1, 1) ), ( (2, 4) ).
- Transforming ( g(x) = (x - 2)^2 )
To get the graph of ( g(x) = (x - 2)^2 ) from ( f(x) = x^2 ), shift the graph of ( f(x) = x^2 ) to the right by ( 2 ) units. All points on the graph of ( f ) can be moved two units to the right.
- Understanding the Equation ( x = |y| ) with ( x = 0 )
For ( x = |y| ) and ( x = 0 ), we have:
$$ 0 = |y| $$
This means ( y = 0 ).
- Graphing ( |x| + |y| = 10 )
The equation represents a diamond shape centered at the origin with vertices at ( (10, 0) ), ( (0, 10) ), ( (-10, 0) ), ( (0, -10) ). This is because both ( |x| ) and ( |y| ) are non-negative.
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The domain of ( y = \frac{2x - 1}{x^2 - x} ) is ( x \in \mathbb{R} ) except ( x = 0, 1 ).
-
The domain of ( f(x) = \sqrt{\frac{1}{x - 1}} ) is ( (1, \infty) ) with a range of ( (0, \infty) ).
-
( f(x) = x^2 ) graphs as a parabola.
-
To obtain ( g(x) = (x - 2)^2 ), shift ( f(x) ) right by 2 units.
-
( x = |y| ) with ( x = 0 ) implies ( y = 0 ).
-
The shape defined by ( |x| + |y| = 10 ) is a diamond centered at the origin with vertices at ( (10, 0) ), ( (0, 10) ), ( (-10, 0) ), ( (0, -10) ).
More Information
The transformations of functions and understanding how to manipulate equations are key parts of studying functions in mathematics. The concept of the absolute value creates unique graphical representations.
Tips
- Failing to consider restrictions on the domain, especially with square roots and denominators.
- Confusing the order of transformations when shifting graphs.
- Not identifying what the absolute value equation represents graphically.
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