Graphing radical functions
1 answer:
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Answer:
0 ≤ Ф ≤ 4π.
Step-by-step explanation:
since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))
the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.
We take
Note that
Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.
The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.
Answer:
y = -4x² + 32x - 48
Step-by-step explanation:
The standard form of a quadratic equation is
y = ax² + bx + c
We must find the equation that passes through the points:
(2, 0), (6,0), and (3, 12)
We can substitute these values and get three equations in three unknowns.
0 = a(2²) + b(2) + c
0 = a(6²) + b(6) + c
12 = a(3²) + b(3) + c
We can simplify these to get the system of equations:
(1) 0 = 4a + 2b + c
(2) 0 = 36a + 6b + c
(3) 12 = 9a + 3b + c
Eliminate c from equations (1) and (2). Subtract (1) from (2).
(4) 0 = 32a + 4b
Eliminate c from equations (2) and (3). Subtract (3) from (2).
(5) -12 = 27a - 3b
Simplify equations (4) and (5).
(6) 0 = 8a + b
(7) -4 = 9a - b
Eliminate b by adding equations (6) and (7).
(8) a = -4
Substitute (4) into (6).
0 = -32 + b
(9) b = 32
Substitute a and b into (1)
0 = 4(-4) + 2(32) + c
0 = -16 + 64 + c
0 = 48 + c
c = -48
The coefficients are
a= -4, b = 32, c = -48
The quadratic equation is
y = -4x² + 32x - 48
The diagram below shows the graph of your quadratic equation and the three points through which it passes.
