# Thread: Transforming √(9 - y²)

1. ## Transforming √(9 - y²)

The relation x = √(9 - y²) is compressed vertically by a factor of (1/3), then translated 1 unit to the right. Determine the equation of the transformed relation.

A. x = √(9 - 9y²) + 1

B. x = √(9 - 9y²) - 1

C. x = √[9 – (y²/9)] + 1

D. x = √[9 – (y²/9)] - 1

This one really beats me. The equation is in the form of an inverse function (unsure if this matters or not) Here is my analysis:

• Vertical compression by a factor of (1/3)
• Horizontal translation, 1 unit to the right

Translation to the right by 1 unit means x=1. Since the function is an inverse function, (A) and (C) is the likely answer since I can see that x = + 1.

Between (A) and (C), I am lost as I cannot see how a vertical compression of (1/3) factors can change the equation to (9 - 9y²) or [9 – (y²/9)].

I was thinking more like √(9 – (1/3)y²) as vertical compression is applied to y-value without taking it’s reciprocal.

2. Hello, shenton!

The relation: .x .= .√(9 - y²) . Note: x > 0
is compressed vertically by a factor of (1/3), then translated 1 unit to the right.
Determine the equation of the transformed relation.
. . . . . . _____
A) x = √9 - 9y² + 1
. . . . . . _____
B) x = √9 - 9y² - 1
. . . . . . ________
C) x = √9 – (y²/9) + 1
. . . . . . ________
D) x = √9 – (y²/9) - 1

I had to baby-talk my way through this one . . .

Square both sides of the equation:. .= .9 - y² . . x² + y² .= .9

We have the right half of a circle centered at the origin with radius 3.

We have a vertical compression by a factor of 1/3.
The result is an ellipse:
. . major axis from (-3,0) to (3,0), minor axis from (-1,0) to (1,0).
Its equation is: .x²/9 + y²/1 .= .1

Translated 1 unit to the right, we'd have: .(x - 1)²/9 + y² .= .1

Solve for x: .(x - 1)²/9 .= .1 - y²

Multiply by 9: .(x - 1)² .= .9 - 9y²
. . . . . . . . . . . . . . . . . . . . .______
Take square root: .x - 1 .= .√9 - 9y²
. . . . . . . . . . . . . . . .______
And we have: . x .= .√9 - 9y² + 1 . . . Answer-choice: A)

3. Thanks for the detailed workings. This is a little harder than I thought.

4. Originally Posted by shenton
The relation x = √(9 - y²) is compressed vertically by a factor of (1/3), then translated 1 unit to the right. Determine the equation of the transformed relation.

A. x = √(9 - 9y²) + 1

B. x = √(9 - 9y²) - 1

C. x = √[9 – (y²/9)] + 1

D. x = √[9 – (y²/9)] - 1

This one really beats me. The equation is in the form of an inverse function (unsure if this matters or not) Here is my analysis:
• Vertical compression by a factor of (1/3)
• Horizontal translation, 1 unit to the right

Translation to the right by 1 unit means x=1. Since the function is an inverse function, (A) and (C) is the likely answer since I can see that x = + 1.

Between (A) and (C), I am lost as I cannot see how a vertical compression of (1/3) factors can change the equation to (9 - 9y²) or [9 – (y²/9)].

I was thinking more like √(9 – (1/3)y²) as vertical compression is applied to y-value without taking it’s reciprocal.

There IS a slightly more direct way.

First when you compress the y axis by a factor of 1/3 you are replacing all y's in the equation with 3y.

When you translate to the right by 1 unit you are replacing all x's in the equation with x - 1.

So:
x = sqrt(9 - y^2)
becomes:
x - 1 = sqrt(9 - (3y)^2)

x = sqrt(9 - 9y^2) + 1

-Dan