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AlekseyPX
2 years ago
10

Consider the function f(x)=x2 bx c. if the minimum of this function is located at point (2,0), then what are the values of b and

c?
Mathematics
1 answer:
galina1969 [7]2 years ago
4 0

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

Read more about functions at:

brainly.com/question/2328150

#SPJ1

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Lena [83]

We are given the expression x³ - 2y² - 3x³ + z⁴. We have to evaluate it at a given set of x, y, and z values. Before then, let's combine like terms.

x³ - 2y² - 3x³ + z⁴

= -2x³ - 2y² + z⁴ <-- by combining like terms

Now we evaluate for x=3 y=5 z = -3

= -2(3³) - 2(5²) + (-3⁴) <-- by putting in for x, y, z

= -2(27) - 2(25) + 81 <--- parentheses and exponents first in order of operations

= -54 - 50 + 81 <---- multiplication is next

= -4 + 81

= 77


Thus, we evaluate and it's 77.

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4 years ago
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3 years ago
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Answer:

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