The values of b and c are -4 and 4 respectively
<h3>How to determine the values of b and c?</h3>
The function is given as:
f(x) = x^2 + bx + c
Differentiate f(x)
f'(x) = 2x + b
Set to 0
2x + b = 0
Solve for b
b = -2x
The minimum is (2, 0).
So, we have:
b = -2 * 2
b = -4
Substitute b = -4 in f(x) = x^2 + bx + c
f(x) = x^2 - 4x + c
Substitute (2, 0)
0 = (2)^2 - 4(2) + c
This gives
0 = 4 - 8 + c
Evaluate
c = 4
Hence, the values of b and c are -4 and 4 respectively
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T= -15 over 2
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Answer:
eigenvalues depends on the characteristic polynomial equation formed A-lambda I
Step-by-step explanation:
Consider that while calculating A--lambda I 2nd order characteristic polynomial is formed , where the characteristic polynomial always has two roots.
SO these are eigen values.
Which can be real imaginary or complex .
In 3X3 the eigen values will be 3.