What is the value of the fourth term in a geometric sequence for which a = 110 and r=0.5?
1 answer:
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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.
