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Special Pythagorean Triple. r 2 = 16 r = 4. r 2 = 25 r = 5. Center @ (-2, 1). Center @ (0, 0). 5. 4. 3. 5. 4. 3. 4. 3. 4. 4. 4. 5. 5. 4. 5. (___, 0). (0, ___). Substitute in 0 for y and solve for x. Substitute in 0 for x and solve for y. ( x , y ).
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Special Pythagorean Triple r2 = 16 r = 4 r2 = 25 r = 5 Center @ (-2, 1) Center @ (0, 0) 5 4 3 5 4 3 4 3 4 4 4 5 5 4 5
(___, 0) (0, ___) Substitute in 0 for y and solve for x. Substitute in 0 for x and solve for y.
( x, y ) ( h, k ) Substitute in h and k as 1 and -2. Substitute in x and y as 4 and 2. Solve for r2. Write the standard equation of a circle with the endpoints of the diameter as ( -2, 7) and ( -14, -9).
GENERAL FORM OF THE EQUATION OF A CIRCLE: Convert to Graph Group x terms and y terms together and move the constant to the other side. by completing the square. Complete the square of the x’s and y’s. (+2)2 (-3)2 4 9 x y x y Center @ (-2, 3) r2 = 4 r = 2 Graph Divide everything by 2. Why? (-3)2 (-2)2 9 4 x y x y Center @ (3, 2) r2 = 25 r = 5
x 0 y a x x y (-a) 2 2 Square both sides to remove radical. FOIL the binomials. Focus Cancel like terms on each side. a 4a 4a a Solve for x2. 2a 2a 2a a -a -a Directrix
Graph the following equations. x = - 3 The y is squared and the coefficient on the x is positive, the parabola opens to the right. 4a = 12, a = 3 and the vertex is at (0, 0). 6 V 3 F 6
Graph the following equations. The x is squared and the coefficient on the y is negative, the parabola opens down. 4a = -16, a = -4 and the vertex is at (0, 0). y = 4 V 4 8 8 F
Graph the following equations. x = 2 The y is squared and the coefficient on the x is negative, the parabola opens to the left. 4a = -8, a = -2 and the vertex is at (0, 0). 4 F V 2 4
Graph the following equations. The x is squared and the coefficient on the y is positive, the parabola opens up. 4a = 8, a = 2 and the vertex is at (2, -1). 4 F 4 2 y = - 3 V
Graph the following equations. x = 3 We need to complete the square of the y-terms to put in graphing form. Isolate the y-terms. (-1)2 1 2 V F 2 Factor out the 4 as the GCF. The y is squared and the coefficient on the x is positive, the parabola opens to the right. 4a = 4, a = 1 and the vertex is at (4, 1).
Graph the following equations. We need to complete the square of the x-terms to put in graphing form. Isolate the x-terms. (+3)2 9 F 2 2 Factor out the 4 as the GCF. y = - 3 V The x is squared and the coefficient on the y is positive, the parabola opens up. 4a = 4, a = 1 and the vertex is at (-3, -2).
Draw a rough graph. Equation format is ... …plug in x & y to solve for 4a. (2,3) Draw a rough graph. Equation format is ... …distance from V to F is 1, a = 1, and plug in the vertex values. V F …distance from F to the directrix line is 6, V is halfway, so a = 3. Plug in a and the vertex values. Draw a rough graph. Equation format is ... F(-4, 4) 3 V(-4, ?) 4 – 3 = 1 V(-4, 1) y = -2