List the vertices of each conic given. Solve the system by identifying all points of intersections.
y = x^2-13
x^2+y^2 = 25
y = x^2-13
x^2+y^2 = 25
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The first is an equation for a parabola. The vertex form is y = a(x-h)² + k, so we read off that the vertex is at (0, -13).
The second equation is that of a circle of radius 5 centered at (0, 0). I don't think a circle has vertices.
Solving the system: solve the second equation for x^2 and substitute that into the first equation:
x^2 + y^2 = 25 ⇒ x^2 = 25 - y^2
y = (25 - y^2) - 13
y^2 + y - 12 = 0
(y + 4)(y - 3) = 0
y = -4 or y = 3
x^2 = 25 - y^2
x = ±√(25 - y^2)
When y = -4, x = ±√(25 - 16) = ±√9 = ±3
When y = 3, x = ±√(25 - 9) = ±√16 = ±4
So there are four points of intersection: (3, -4), (-3, -4), (4, 3), and (-4, 3)
The second equation is that of a circle of radius 5 centered at (0, 0). I don't think a circle has vertices.
Solving the system: solve the second equation for x^2 and substitute that into the first equation:
x^2 + y^2 = 25 ⇒ x^2 = 25 - y^2
y = (25 - y^2) - 13
y^2 + y - 12 = 0
(y + 4)(y - 3) = 0
y = -4 or y = 3
x^2 = 25 - y^2
x = ±√(25 - y^2)
When y = -4, x = ±√(25 - 16) = ±√9 = ±3
When y = 3, x = ±√(25 - 9) = ±√16 = ±4
So there are four points of intersection: (3, -4), (-3, -4), (4, 3), and (-4, 3)