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.
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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