# Tutor profile: Sophie E.

## Questions

### Subject: The Adobe Suite

How do I create a parent layer in After Effects?

On the layer the you want to make a child, you should see a button that has a swirl on it. Click on the swirl and drag it to the layer that you want to be the parent. An arm should extend from the child layer to the parent layer and latch onto the parent layer. You have now created a parent layer.

### Subject: Pre-Calculus

Find the center and radius of the circle. x^2 + 16x + y^2 - 6y - 8 = 0

In order to find the center and the radius, we must put the equation into center-radius form. That means making the equation fit into this form: (x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius. To do this we need to complete the square. First, move the 8 to the other side. x^2 + 16x + y^2 - 6y = 8 Then put all x terms into one parenthesis and all y terms into another. (x^2 + 16x) + (y^2 - 6y) = 8 Next, we must add (b/2)^2 to each grouping. Our goal is to end up with a perfect square, two-degree polynomial for each grouping. Right now, we have both groups in the form ax^2 + bx + c, but both groups are missing a "c" term. That is where the (b/2)^2 comes into play: that will give us the term for each group that will stand in for c. For the "x" grouping, b is 16, so c is (16/2)^2 = 64 For the "y" grouping, b is -6, so c is (-6/2)^2 = 9 So we add a 64 to the x group and a 9 to the y group. But don't forget that whatever we add to one side, we must add to the other side. (x^2 + 16x + 64) + (y^2 - 6y + 9) = 8 + 64 + 9 (x^2 + 16x + 64) + (y^2 - 6y + 9) = 81 Now, we just factor both groups. Since they are both perfect squares, we can make the equation: (x + 8)^2 + (y - 3)^2 = 81 Comparing that to the center-radius form of the circle equation we can see that the center, or (h,k) is: (-8, 3) and the radius is: 9

### Subject: Algebra

Solve. 2x^3 - 17x^2 - 5x - 14 = -2x^3 + 11x^2 - 7x

Our first step is to move all terms to one side of the equation. This will leave zero on one side and allow us to factor. We can move terms by performing the opposite action to both sides. 2x^3 - 17x^2 - 5x -14 = -2x^3 + 11x^2 - 7x +2x^3 -11x^2 +7x +2x^3 -11x^2 +7x This creates: 4x^3 - 28x^2 + 2x -14 = 0 From here, we can begin to factor. There are a few ways to factor a third degree polynomial: grouping, sum of cubes, difference of cubes, and synthetic division. Typically when there are four terms as there are here, you should start by trying to factor by grouping. That is done by grouping the first two and last two terms using parenthesis. (4x^3 - 28x^2) + (2x -14) = 0 Now, we factor out any common factors from each group. The first group of terms share a common 4x^2 and the second group of terms share a 2. 4x^2 (x - 7) + 2 (x - 7) = 0 From that action, we can see that (x - 7) is shared among both groups. Because of that, we can factor out the (x - 7) as such: (x - 7)(4x^2 + 2) = 0 Now we have factored the polynomial, so we just have to solve. That is done so by setting each group equal to 0. x - 7 = 0 x = 7 4x^2 + 2 = 0 -2 -2 4x^2 = -2 ÷ 4 ÷ 4 x^2 = -1/2 Take the square root of both sides. x = √(-1/2) x = ± i (√(1/2)) and x = 7

## Contact tutor

needs and Sophie will reply soon.