advertisement

Aim: Using intersecting curves to solve algebra problems. Do Now: Solve for all values of x. 2. y=√8 (x + 3)2 + y2 = 9 (x + 3)2 + 8= 9 Substitute in y x2+ 6x + 9 +8 = 9 Factor out x2 + 6x +8= 0 Simplify (x + 4) (x + 2) = 0 x= -4, -2 3. y=x2 x2 + (y-7)2=13 y + (y-7)2 = 13 y + y2-14y+49=13 y2- 13y+36=0 (y-9)(y-4)=0 y=9 or y=4 x= ±3 x= ± 2 4. y=x2 (x+2)2+ (y-2)2=36 (x+2)2+ (x2-7)2=36 x2+ 4x+ 4+ x4-14x+13=0 x4-13x2+ 4x+17=0 y=√8 (x+3)2+ y2=9 Intersection of circle with center (-3,0) r=3 and line y=√8 x2+6x+8=0 *If you were to graph this to find the solution for this quadratic equation, instead of graphing the parabola you can graph a simpler circle and horizontal line. ● ● Horizontal line intersects circle at 2 points. Only for equations x2+bx+c=0 with c > 0 Line would be : y= √c b b Circle would be: (x + 2 )2 + y2= (2)2 Ex. x2 + 6x+ 8 =0 y=√8 (x+3)2 + y2= 9 Parabolas and circles can intersect to up four places. ● ● 2 pts. 3 pts. 4 pts. With do now # 4, instead of doing it algebraically, you can solve it graphically by graphing a circle and intersecting it with the parabola y=x2. Do now #4: x4 – 13x2 + 4x + 17= 0 ( it should have 4 ans.) depressed quartic= 4th degree equation with no x3 term. Take -13x2 and change it to x2-14x2 x4+ x2-14x2 + 4x +17=0 Change x4 to y2 and 14x2 turns to 14y2 and simplify (because you substitute it from the parabola y=x2) y2+ x2-14y +4x = -17 Bring y terms together and complete the square y2-14y+ 49 + x2+4x+4= -17 + 49 +4 This becomes the circle you need : (y-7)2 + (x+2)2 =36 This circle with center (7,-2) and radius 6 will intersect with parabola y=x2. The intersections would be the solution set for the original problem: x4-13x2+4x+17=0