Writing Quadratic Equations when Given Vertex and Focus/ Directrix. Vanessa Ledford Algebra ll

Definition of Focus: The parabola focus is a point from which distances are measured in forming a parabola and where these distances converge. After changing the quadratic functions to vertex form, the formula for the Focus is (h, k+¼a) “a” is the number that is in front of the parenthesis. Vertex: (-1,-4) Axis of symmetry: x = -1 Focus: (h, k+1/4a) (-1,-4+¼( 1 )) -4+ (-0.25) = (-1,- 3.75) f(x) = (x+1)2 – 4

Definition of Directrix: A parabola Directrix is a line from which distances are measured in forming a parabola. After changing the quadratic function in vertex form the formula for the Directrix is y = k – 1/4a Vertex: (-1,-4) Directrix: y = k – 1/4a y = (-4) – ¼(1) y = -4.25 f(x)= (x+1)2 – 4

Definition of the Axis of Symmetry: The line perpendicular to the directrix and passing through the focus is called the axis of symmetry.

Steps to use: 1- Make a graph 2- Find the Vertex 3- Convert the quadratic function into Vertex Form 4- Next plug in the numbers for the focus or directrix while using the equation that applies to each of them. Ex: If the focus is given then use (h, k+¼a), or if given the directrix use y = k – 1/4a. In certain cases if your given both the focus and directrix it doesn’t matter which equation to use. 5- Find the equation to use for finding the leading coefficient -“a” 6- Then solve for “a” 7- Finally write the Vertex Form

Showing a problem using formula for Focus: Write the equation of the quadratic function with a vertex at (0,1) and focus at (0,4). 2. Vertex = (0,1) 6. 1+ (1/4a) = 4 -1 -1 (1/4a) = 3 (4a/1) * (1/4a) = 3 * (4a/1) 1 = 12a (1/12)= (12a)/12 a = 1/12 1. 3. f(x) = a(x – 0)2 +1 a(x)2 +1 4. (h, k + 1/4a) = (0,4) (0, 1 + 1/4a) = (0,4) 5. 1+ 1/4a = 4 7. f(x) = 1/12(x)2 + 1

Showing a problem using formula for Directrix: Write the equation of the quadratic function with vertex at (1,8) and directrix at y = 3. 2. Vertex = (1, 8) 6. 3 = 8- (1/4)a -8 -8 -5 = (-1/4a) (-4a/1) *-5 = (-1/4a) * (-4a/1) 1 = 20a 1/20= 20a/20 a = 1/20 1. 3. f(x) = a(x-1)2 + 8 4. y = k – 1/4a y = 3 5. 3 = 8- 1/4a 7. f(x) = 1/20 (x -1)2 +8

Example 1: Write the equation of the quadratic function with a vertex at (3, -5) and focus at (3, -4.25). 6. -5 + 91/4a) = -4.25 +5 +5 (1/4a) = 0.75 (4a/1)* (1/4a) = 0.75 * (4a/1) 1= 3a (1/3) = (3a)/3 a= 1/3 1. 2. Vertex = (3, -5) 3. f(x) = a (x- 3)2 -5 4. (h, k + 1/4a) = (3, -4.25) ( 3, -5 + 1/4a) = (3, -4.25) 5. -5 + 1/4a = -4.25 7. f(x) = 1/3 (x-3)2 – 5

Example 2: 7. f(x) = -1/16(x – 4)2 + 6 Write the equation of the quadratic function with focus at (4, 2) and directrix at y = 10. 8/2 = 4 6. 6 + 1/4a = 2 -6 -6 (1/4a) = -4 4a/1 * (1/4a) = -4 * 4a/1 1 = -16a (-1/16) = -16a/-16 a = -1/16 1. 2. Vertex = (4, 6) 3. f(x) = a (x – 1/4a) = (4,2) 4. (h, k = 1/4a) = (4,2) (4, 6 + 1/4a) = (4,2) 5. 6 + 1/4a = 2 7. f(x) = -1/16(x – 4)2 + 6

Example 3: Write the equation of the quadratic function with focus at (6, 5) and directrix at y = -2. 7/-2 = -3.5 6. -2 = 1.5 – 1/4a -1.5 -1.5 -3.5 = -1/4a (-4a/1) * -3.5 = (-1/4a) * (-4a/1) 1 = 14a (1/14) = (14a)/14 a = 1/14 1. 2. Vertex = (6, 1.5) 3. f(x) = a (x – 6)2 +1.5 4. y = k – 1/4a y = -2 5. -2 = 1.5 – 1/4a 7. f(x) = 1/14(x – 6)2 +1.5

Work Cited 3.6 Focus and Directrix – Writing Quadratics from Info Part ll Home Work. 3.6 Focus and Directrix http://www.purplemath.com/modules/conics.htm