1. ## Test Review

Have a couple more questions I'm struggling to understand for tomorrows test, if anyone can reply and explain to me how I reach my answer exactly or provide me a link where I can learn how to conquer the equation that would be greatly appreciated.
Question 1:
Consider the circle given X^2+Y^2=1. Determine the equation of the transformed function if the graph undergoes a horizontal stretch by a factor of 1/2, vertical stretch by a factor of 3, horizontal translation of 1 unit right and a vertical translation of 4 units down. (I essentially don't know where to start here)

Question 2:
Consider the conic section given by (x-2)^2 + (y+3)^2. Convert to general form and state the coordinates of the center and vertices.
36 16

Question 3:
AngleƟ has a terminal arm in the third quadrant. If cotƟ= 12, the values of the remaining five primary and reciprocal trig ratios.
5

Any help is greatly appreciate!
Thanks,

Shane

2. ## Re: Test Review

Originally Posted by Prentz
Question 1:
Consider the circle given X^2+Y^2=1. Determine the equation of the transformed function if the graph undergoes a horizontal stretch by a factor of 1/2, vertical stretch by a factor of 3, horizontal translation of 1 unit right and a vertical translation of 4 units down.
Consider a horizontal stretch by a factor of 1/2. Then each point (x, y) on the plot becomes (x/2, y). Let's call this new point (x', y'); then x' = x/2 and y' = y, or x = 2x' and y = y'. Substituting x(x') and y(y') into the original equation, we get (2x')^2 + (y')^2 = 1, which is the new equation. The remaining steps can be done similarly.

Originally Posted by Prentz
Question 2:
Consider the conic section given by (x-2)^2 + (y+3)^2. Convert to general form and state the coordinates of the center and vertices.
This is not an equation. However, for any R, (x-2)^2 + (y+3)^2 = R^2 is a circle with center (2, -3) and radius R.

Originally Posted by Prentz
Question 3:
AngleƟ has a terminal arm in the third quadrant. If cotƟ= 12, the values of the remaining five primary and reciprocal trig ratios.
Since a terminal arm is in the third quadrant, sin Ɵ and cos Ɵ are negative. Denoting x = -cos Ɵ and y = -sin Ɵ, you can write a relation between x and y and then use the Pythagorean theorem to find x and y.

3. ## Re: Test Review

Thank you Emakarov, to clear it up and if this helps the first question 2 was suppose to be:
[U](x-2)^2/36 + [U](y+3)^2/16
for some reason it didn't allow the number to stay under the equation for the division.
Also Question 3 is: cotƟ=12/5
again the equation didn't have the 5 under it.
If this changes anything please let me know!

4. ## Re: Test Review

[QUOTE=Prentz;711136]the first question 2 was suppose to be:
[U](x-2)^2/36 + [U](y+3)^2/16[quote]My guess is this means $\frac{(x-2)^2}{36}+\frac{(y+3)^2}{16}$. This is still not an equation, but $\left(\frac{x-2}{6}\right)^2+\left(\frac{y+3}{4}\right)^2=1$ corresponds to an ellipse with the center at (2, -3).

Originally Posted by Prentz
Convert to general form and state the coordinates of the center and vertices.
By "vertices" do you mean foci (plural of "focus")? They are on the major axes at the distance $\sqrt{a^2-b^2}$ from the center (a and b are semi-major axis and semi-minor axis, respectively).

Originally Posted by Prentz
Also Question 3 is: cotƟ=12/5
This does not change the solution method.

5. ## Re: Test Review

Thank you for the help I have sorta out question 1 and 3 now and as far as question 2 yes I mean find the center and foci need to be found.

6. ## Re: Test Review

Originally Posted by Prentz
as far as question 2 yes I mean find the center and foci need to be found.
I am not sure if this presents a difficulty to you in view of post #4.

7. ## Re: Test Review

[QUOTE=emakarov;711139][QUOTE=Prentz;711136]the first question 2 was suppose to be:
[U](x-2)^2/36 + [U](y+3)^2/16
My guess is this means $\frac{(x-2)^2}{36}+\frac{(y+3)^2}{16}$. This is still not an equation, but $\left(\frac{x-2}{6}\right)^2+\left(\frac{y+3}{4}\right)^2=1$ corresponds to an ellipse with the center at (2, -3).

By "vertices" do you mean foci (plural of "focus")? They are on the major axes at the distance $\sqrt{a^2-b^2}$ from the center (a and b are semi-major axis and semi-minor axis, respectively).
The vertices of an ellipse are not the foci, they are the points on the ellipse where the curvature is maximum or minimum. They are the points at the ends of the axes. For an ellipse of the form $\frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1$ they are at $(x_0, y_0+ b)$, $(x_0, y_0- b)$, $(x_0+ a, y_0)]$, and [tex]x_0- a, y_0)[]/tex].

This does not change the solution method.

8. ## Re: Test Review

Thank you halls I appreciate you clearing that up, Is it possible to have you apply that to the equation I posted above? Just to give me a better understand of how I complete the question?