Graphing a function and finding the domain and range

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SUMMARY

The discussion focuses on graphing the function y = -4(√(4x - 1)) - 1 and determining its domain and range. The domain is established as x ≥ 1/4, derived from solving the inequality 4x - 1 ≥ 0. The participants clarify that the transformations involve shifting the graph down by 1 unit and reflecting it vertically, while the factor of -4 expands the function vertically. A step-by-step approach to graphing is recommended, starting from y = √x and applying transformations sequentially.

PREREQUISITES
  • Understanding of function transformations, including vertical shifts and reflections.
  • Knowledge of inequalities and how to solve them, specifically for determining the domain.
  • Familiarity with the basic graph of y = √x.
  • Ability to interpret and apply mathematical notation and expressions accurately.
NEXT STEPS
  • Learn about function transformations in detail, focusing on vertical and horizontal shifts.
  • Study how to graph composite functions and their transformations systematically.
  • Explore the properties of square root functions and their domains.
  • Practice solving inequalities to determine the domain of various functions.
USEFUL FOR

Students studying algebra, mathematics educators, and anyone looking to improve their skills in graphing functions and understanding domain and range concepts.

land_of_ice
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Homework Statement


y= -4(√4x-1) -1
in case it's not clear, the 4x-1 part is all under the radical
the question is to graph and find the domain and range

Homework Equations


The Attempt at a Solution


This is probably what your supposed to do , but not sure, OK
To get the domain set everything under the radical to zero and solve for x, x is 1/4 so that's the domain, but how to get the range??
To graph this, first graph the basic graph of y = √
Then, does it matter what order you do the other things in ?
Some of the other things to do apparently, are:
With regards to the -1 at very right side of the problem means to move the graph to the right along the x axis?
The -4 means to reflect about the y-axis and shrink by 4 ?

Confused - There are more steps to this or different steps or in different order right?
 
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land_of_ice said:

Homework Statement


y= -4(√4x-1) -1
in case it's not clear, the 4x-1 part is all under the radical
the question is to graph and find the domain and range

Homework Equations


The Attempt at a Solution


This is probably what your supposed to do , but not sure, OK
To get the domain set everything under the radical to zero and solve for x, x is 1/4 so that's the domain, but how to get the range??
Not quite. The domain consists of all real values of x for which the function is defined. You've only specified one point. When you solve the equation 4x-1=0, you're finding the boundary between where y(x) is defined and not defined. You still need to figure out which side is which. The function is defined only when 4x-1≥0. If you solve the inequality for x, you get x≥1/4. That's the domain.
To graph this, first graph the basic graph of y = √
Then, does it matter what order you do the other things in ?
Yes, the order matters, but there's not just one ordering that works. You have some leeway here.
Some of the other things to do apparently, are:
With regards to the -1 at very right side of the problem means to move the graph to the right along the x axis?
No, it shifts the function down by 1 in the vertical direction.
The -4 means to reflect about the y-axis and shrink by 4?
The negative sign does indicate a reflection, but the factor of 4 expands, not shrinks, the function in the vertical direction.
Confused - There are more steps to this or different steps or in different order right?
Yes. For one thing, you haven't accounted for what's under the radical yet.

Start from the graph of [tex]y=\sqrt{x}[/tex] and do one thing at a time until you get to the original function. At each step, figure out what happens to the graph. For instance, when you go from

[tex]y=\sqrt{x}[/tex]

to

[tex]y=\sqrt{4x}[/tex],

you compress the graph in the horizontal direction. Now figure out what happens to the new graph when you do another transformation. Keep adding transformations until you get to the desired function. For instance, the following progression will get you to the desired graph:

[tex]\begin{align*}<br /> y & = \sqrt{4(x-1/4)} = \sqrt{4x-1} \\<br /> y & = 4\sqrt{4x-1} \\<br /> y & = -4 \sqrt{4x-1} \\<br /> y & = -4 \sqrt{4x-1}-1<br /> \end{align*}[/tex]

Explain what happens at each step.
 
vela said:
Not quite. The domain consists of all real values of x for which the function is defined. You've only specified one point. When you solve the equation 4x-1=0, you're finding the boundary between where y(x) is defined and not defined. You still need to figure out which side is which. The function is defined only when 4x-1≥0. If you solve the inequality for x, you get x≥1/4. That's the domain.

Yes, the order matters, but there's not just one ordering that works. You have some leeway here.

No, it shifts the function down by 1 in the vertical direction.

The negative sign does indicate a reflection, but the factor of 4 expands, not shrinks, the function in the vertical direction.

Yes. For one thing, you haven't accounted for what's under the radical yet.

Start from the graph of [tex]y=\sqrt{x}[/tex] and do one thing at a time until you get to the original function. At each step, figure out what happens to the graph. For instance, when you go from

[tex]y=\sqrt{x}[/tex]

to

[tex]y=\sqrt{4x}[/tex],

you compress the graph in the horizontal direction. Now figure out what happens to the new graph when you do another transformation. Keep adding transformations until you get to the desired function. For instance, the following progression will get you to the desired graph:

[tex]\begin{align*}<br /> y & = \sqrt{4(x-1/4)} = \sqrt{4x-1} \\<br /> y & = 4\sqrt{4x-1} \\<br /> y & = -4 \sqrt{4x-1} \\<br /> y & = -4 \sqrt{4x-1}-1<br /> \end{align*}[/tex]

Explain what happens at each step.

Thanks, you answered every little thing that was important about the question - Excellent :) Once again, thanks :) :)
 

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