Homework Answers 1) vertex: (1, 2) focus: (1, 4) Axis of Sym: x = 1 directrix: y = 0 2) Eqn: (y + 3)2 = -12(x – 1) vertex: (1, -3) focus: (-2, -3) Axis of Symm: y = -3 directrix: x = 4 3) Eqn: (x + 4)2 = -4(y – 2) vertex: (-4, 2) focus: (-4, 1) A of S: x = - 4 directrix: y = 3 6) x2 = 8y 7) 4 in

PG. 14 HW #19: Write the equation for the given parabola. Section 7.1 Notes, continued PG. 14 HW #19: Write the equation for the given parabola. We can see that the orientation (direction it opens) is: Think about the parts of the equation we need for a parabola opening down: (but p must be negative since it opens down) We need to find the vertex (h,k) and p. The vertex is (3, 5). p always represents the distance from the vertex to the focus, so p = Plug h, k, and p into the standard form of the equation: (x – 3)2 = 4(–1)(y – 5) and simiplify: (x – 3)2 = –4(y – 5) DOWN -1

A) focus (2, 1) and vertex (−5, 1) Example 4: Write an equation for and graph a parabola with the given characteristics. A) focus (2, 1) and vertex (−5, 1) PLOT the INFORMATION to determine the orientation: Think about the parts of the equation we need for a parabola opening right: (but p must be positive since it opens right) We need to find the vertex (h,k) and p. The vertex is (-5, 1). p always represents the distance from the vertex to the focus, so p = Plug h, k, and p into the standard form of the equation: (y – 1)2 = 4(7)(x – -5) [be careful about the location of h and k] and simiplify: (y – 1)2 = 28(x + 5) RIGHT 7 V F

B) vertex (3, −2), directrix y = −1 Example 4: Write an equation for and graph a parabola with the given characteristics. B) vertex (3, −2), directrix y = −1  PLOT the INFORMATION to determine the orientation: Think about the parts of the equation we need for a parabola opening down: (but p must be negative since it opens down) We need to find the vertex (h,k) and p. The vertex is (3, -2). p always represents the distance from the vertex to the focus, so p = Plug h, k, and p into the standard form of the equation: (x – 3)2 = 4(–3)(y – -2) and simiplify: (x – 3)2 = –12(y + 2) DOWN –3 V

C) focus (−1, 7), opens up, contains (3, 7) Example 4: Write an equation for and graph a parabola with the given characteristics. C) focus (−1, 7), opens up, contains (3, 7)   F R PLOT the INFORMATION Think about the parts of the equation we need for a parabola opening up: (p must be positive since it opens up) We need to find the vertex (h,k) and p. The vertex always aligns with the focus, so we know (-1, k). The h value is -1. To find the k value, use the ordered pair pattern for the FOCUS on the blue equation sheet. Focus: (h, k+p) = (-1, 7). If h = -1, then k+p = 7. Solve for k. k = 7 – p This makes the vertex (-1, 7-p). We now have an h and k that we can plug in, as well as an x-value and y-value (3, 7) to plug into the standard form. But which vertex?

C) focus (−1, 7), opens up, contains (3, 7) Example 4: Write an equation for and graph a parabola with the given characteristics. C) focus (−1, 7), opens up, contains (3, 7)   F What we know: vertex (-1, 7-p); point on the circle: (3, 7) as (x, y) Plug h, k, x, and y into standard form of the equation to solve for p. (3 – -1)2 = 4p(7 – (7-p)) (4)2 = 4p(p) 16 = 4p2 4 = p2 p = 2, but since we know it opens up, p = 2 Since p = 2, find the k-value of the vertex using previous steps: Focus: (h, k+p) = (-1, 7), where k+p=7, and k = 7 – p k = 7 – 2 , Plug h, k, and p into the standard form of the equation: (x – -1)2 = 4(2)(y – 5) and simiplify: k = 5 makes the vertex (-1, 5) (x + 1)2 = 8( y – 5)