Guys I am literally lost, I need help please

[Rose34]

Sketch the graph of the following relation, stating the implied domain and range?

(2x−1)^2 + (2y−4)^2 = 1

[dream chaser]

Sketch the graph of the following relation, stating the implied domain and range?

(2x−1)^2 + (2y−4)^2 = 1

Hi Rose34,

Before sketching this equation on a graph, it is important to identify what kind of relation this is. Since the equation follows that of:     
x² + y² = r², the relation is therefore a circle.

Now, since you know that, you would want to manipulate the equation so you know the transformations that are occurring(i.e make the equation into the form (x±a)² +(y±b)² = r²) and thus, be able to draw the circle and be able to state its implied domain and range.

This is how I would do this:

Step 1: You want to expand the equation
(2x−1)² + (2y−4)² =1
(4x²-4x+1) + (4y²-16y+16) = 1

Step 2: Get the common factor out of each bracket to make the coefficient of x²  and y² one(i.e 1)
4(x²-x+1/4) + 4(y²-4y+4)=1

Step 3: Factorize both brackets now
4(x-1/2)² + 4(y-2)² =1

Step 4: Again, like in Step 4, get the common factor out
4[(x-1/2)² + (y-2)²] = 1

Step 5: Move the common factor to the other side of the equal sign(i.e the right hand side of the equal sign)
(x-1/2)² + (y-2)² = 1/4

Step 6: Make the equation into the form of  (x±a)² +(y±b)² = r²
(x-1/2)² + (y-2)² = (√ 1/4)²
(x-1/2)² + (y-2)² = (1/2)²

Now from this you know that the center of the circle is located at (1/2,2) while the radius of the circle is 1/2.

To find the Implied Domain and Range of this relation, all you need to do now is ± the radius(i.e 1/2) from the center of the circle, as this follows the geometrical properties of a circle.

Implied Domain:
1/2 - 1/2 = 0
1/2 + 1/2 = 1

Therefore the Implied Domain is x∈ [0,1]

Implied Range:
2 - 1/2 = 3/2
2 + 1/2 = 5/2

Therefore the Implied Range is y∈ [3/2,5/2]

A graph of the relation will be shown in the attachment of this post.

So that is basically how you answer this question Rose34. Apologies if I have made a mistake somewhere or if anything is unclear. Let me know if there is anything you are still confused about, whether it is the question itself, still unsure how to answer the question or my working out. I hope I have helped you understand how to do this question now.

Thanks,
dream chaser 

[Rose34]

Hi Rose34,

Before sketching this equation on a graph, it is important to identify what kind of relation this is. Since the equation follows that of:     
x² + y² = r², the relation is therefore a circle.

Now, since you know that, you would want to manipulate the equation so you know the transformations that are occurring(i.e make the equation into the form (x±a)² +(y±b)² = r²) and thus, be able to draw the circle and be able to state its implied domain and range.

This is how I would do this:

Step 1: You want to expand the equation
(2x−1)² + (2y−4)² =1
(4x²-4x+1) + (4y²-16y+16) = 1

Step 2: Get the common factor out of each bracket to make the coefficient of x²  and y² one(i.e 1)
4(x²-x+1/4) + 4(y²-4y+4)=1

Step 3: Factorize both brackets now
4(x-1/2)² + 4(y-2)² =1

Step 4: Again, like in Step 4, get the common factor out
4[(x-1/2)² + (y-2)²] = 1

Step 5: Move the common factor to the other side of the equal sign(i.e the right hand side of the equal sign)
(x-1/2)² + (y-2)² = 1/4

Step 6: Make the equation into the form of  (x±a)² +(y±b)² = r²
(x-1/2)² + (y-2)² = (√ 1/4)²
(x-1/2)² + (y-2)² = (1/2)²

Now from this you know that the center of the circle is located at (1/2,2) while the radius of the circle is 1/2.

To find the Implied Domain and Range of this relation, all you need to do now is ± the radius(i.e 1/2) from the center of the circle, as this follows the geometrical properties of a circle.

Implied Domain:
1/2 - 1/2 = 0
1/2 + 1/2 = 1

Therefore the Implied Domain is x∈ [0,1]

Implied Range:
2 - 1/2 = 3/2
2 + 1/2 = 5/2

Therefore the Implied Range is y∈ [3/2,5/2]

A graph of the relation will be shown in the attachment of this post.

So that is basically how you answer this question Rose34. Apologies if I have made a mistake somewhere or if anything is unclear. Let me know if there is anything you are still confused about, whether it is the question itself, still unsure how to answer the question or my working out. I hope I have helped you understand how to do this question now.

Thanks,
dream chaser

Thank you so much!!
You did no mistake at all and your explanation is very clear and good. By the way, are you doing methods as well?

[dream chaser]

Thank you so much!!
You did no mistake at all and your explanation is very clear and good. By the way, are you doing methods as well?

Hi Rose34,

Thanks for the feedback. It is very much appreciated.

No, I finished VCE last year and have just started Uni this year. I did Methods in Year 11 by the way.

Let me know if you get stuck with any other questions.

Thanks,
dream chaser